Chapter 10 Circles (Additional Questions)

The definition given in the question matches the definition of a circle.

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A circle is the locus of a point in a plane such that its distance from a fixed point (the center) is constant (the radius).

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Therefore, the collection of all points in a plane, which are at a fixed distance from a fixed point in the plane, is called a Circle.

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The correct option is (B) Circle.

In the definition of a circle, "the collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane", the fixed point is the central point from which all points on the circle are equidistant.

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This fixed point is known as the center of the circle.

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The fixed distance is called the radius.

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Therefore, the fixed point in the definition of a circle is called the Center.

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The correct option is (C) Center.

A chord of a circle is defined as a straight line segment whose endpoints both lie on the circle.

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Let's examine the given options:

(A) The center to any point on the circle describes a radius.

(B) Any two points on the circle correctly describes a chord.

(C) The center to two points on the circle does not describe a standard geometric term.

(D) Two points outside the circle describes a line segment that does not intersect the circle or is a segment of a secant line, but not a chord.

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Thus, a chord of a circle is a line segment joining any two points on the circle.

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The correct option is (B) Any two points on the circle.

A chord is a line segment connecting two points on a circle. The length of a chord varies depending on its position in the circle.

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The diameter of a circle is a special type of chord that passes through the center of the circle.

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Consider any chord of the circle. The distance of the chord from the center is inversely proportional to its length. Chords farther from the center are shorter, while chords closer to the center are longer.

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The chord that is closest to the center (passing through the center) has the maximum possible length. This chord is the diameter.

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Therefore, the longest chord of a circle is its diameter.

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The correct option is (B) Diameter.

In a circle, a chord divides the circle into two regions. Each region is bounded by the chord and the arc connecting the endpoints of the chord.

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This region is known as a segment of the circle.

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Let's look at the other options:

(A) A sector is the region bounded by two radii and the arc between them.

(C) A quadrant is a quarter of a circle, bounded by two perpendicular radii and the arc between them.

(D) A chord is the line segment itself, not the region.

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Therefore, the region between a chord and the corresponding arc is called a segment.

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The correct option is (B) Segment.

Two geometric figures are said to be congruent if they have the same shape and the same size. For circles, the shape is always the same (a perfect round form).

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The size of a circle is determined solely by its radius. If two circles have the same radius, they are identical in size and shape, and thus congruent.

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Let's examine the other options:

If two circles have equal Areas, say $A_1 = A_2$. The area of a circle is given by the formula $A = \pi r^2$. So, $\pi r_1^2 = \pi r_2^2$. Dividing by $\pi$ (which is non-zero), we get $r_1^2 = r_2^2$. Since the radius must be a positive value, this implies $r_1 = r_2$. Thus, equal areas imply equal radii.

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If two circles have equal Circumferences, say $C_1 = C_2$. The circumference of a circle is given by the formula $C = 2\pi r$. So, $2\pi r_1 = 2\pi r_2$. Dividing by $2\pi$ (which is non-zero), we get $r_1 = r_2$. Thus, equal circumferences imply equal radii.

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Since equal areas, equal radii, and equal circumferences are all equivalent conditions for circles (each one implies the others) and each guarantees the circles are of the same size, any of these conditions is sufficient for two circles to be congruent.

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Therefore, the most comprehensive correct option is that two circles are congruent if and only if they have equal All of the above.

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The correct option is (D) All of the above.

This question refers to a key theorem in circle geometry.

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The theorem states that the angle subtended by an arc (or the corresponding chord) at the center is twice the angle subtended by the same arc (or chord) at any point on the remaining part of the circle (the circumference).

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Let the angle subtended by the chord at the center be $\theta_c$ and the angle subtended by the same chord at a point on the remaining part of the circle be $\theta_a$. According to the theorem, we have the relationship:

This means the angle at the center is double the angle at the circumference (on the remaining part).

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Therefore, the angle subtended by a chord at the center is double of the angle subtended by the same chord at any point on the remaining part of the circle.

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The correct option is (C) Double of.

Given:

Angle subtended by a chord at the center = $100^\circ$.

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To Find:

The angle subtended by the same chord at a point on the major arc.

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Solution:

We use the theorem which states that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

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Let the chord be AB. The angle subtended by the arc AB at the center O is $\angle \text{AOB} = 100^\circ$.

Let C be a point on the major arc AB. The angle subtended by the same arc (minor arc AB) at point C on the remaining part of the circle is $\angle \text{ACB}$.

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According to the theorem, we have:

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Substitute the given value:

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Solving for $\angle \text{ACB}$:

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Thus, the angle subtended by the same chord at a point on the major arc is $50^\circ$.

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The correct option is (B) $50^\circ$.

This question is based on a fundamental theorem in circle geometry.

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The theorem states that equal chords of a circle (or of congruent circles) are equidistant from the center.

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Equidistant means that the perpendicular distance from the center to each of the equal chords is the same.

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Therefore, equal chords of a circle are equidistant from the center.

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The correct option is (C) Center.

An angle in a semicircle is defined as the angle subtended by the diameter at any point on the circumference of the circle.

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According to a fundamental theorem in geometry, the angle subtended by a diameter at any point on the circumference is always a right angle.

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A right angle measures $90^\circ$.

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Therefore, the angle in a semicircle is always a right angle.

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The correct option is (C) Right.

This question relates to the properties of chords and arcs in a circle.

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A theorem in circle geometry states that if two chords of a circle (or of congruent circles) are equal, then their corresponding arcs (both minor and major) are congruent.

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Conversely, if two arcs of a circle (or of congruent circles) are congruent, then their corresponding chords are equal.

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Given that the two chords of a circle are equal, it directly follows from this theorem that their corresponding arcs must be congruent.

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Therefore, if two chords of a circle are equal, then their corresponding arcs are congruent.

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The correct option is (B) Congruent.

A cyclic quadrilateral is a quadrilateral whose vertices all lie on a single circle.

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The boundary of a circle is known as its circumference.

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For a quadrilateral to be cyclic, all four of its vertices must lie on this boundary.

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Let's examine the options:

(A) If the vertices lie in the Interior, they are inside the circle but not necessarily on the boundary.

(B) If the vertices lie in the Exterior, they are outside the circle.

(C) If the vertices lie on the Circumference, they are on the boundary of the circle, which is the definition of a cyclic quadrilateral.

(D) The Center is a single point, not where the vertices of a quadrilateral would lie.

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Therefore, a quadrilateral ABCD is cyclic if all four vertices A, B, C, and D lie on the Circumference of a circle.

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The correct option is (C) Circumference.

A cyclic quadrilateral is a quadrilateral whose vertices lie on a circle.

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A fundamental property of cyclic quadrilaterals is that the sum of the measures of opposite angles is supplementary, which means their sum is $180^\circ$.

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If ABCD is a cyclic quadrilateral, then:

and

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Therefore, in a cyclic quadrilateral, the sum of a pair of opposite angles is $180^\circ$.

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The correct option is (B) $180^\circ$.

This question is the converse of the property of a cyclic quadrilateral.

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The theorem states that if the sum of a pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic. This means its vertices lie on a circle.

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Let's consider the other options:

(A) A parallelogram has opposite angles equal, not necessarily supplementary (unless it's a rectangle or a square, which are special cases of cyclic quadrilaterals).

(B) A rhombus has opposite angles equal, not necessarily supplementary (unless it's a square).

(D) A trapezium (trapezoid) is a quadrilateral with at least one pair of parallel sides. The sum of consecutive interior angles between parallel sides is $180^\circ$, but the sum of opposite angles is not necessarily $180^\circ$ (unless it's an isosceles trapezium, which is cyclic).

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The defining condition for a quadrilateral to be cyclic is that the sum of its opposite angles is $180^\circ$.

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Therefore, if the sum of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic.

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The correct option is (C) Cyclic.

Assertion (A): The perpendicular from the center of a circle to a chord bisects the chord.

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This is a fundamental theorem in circle geometry. It states that if a line segment is drawn from the center of a circle perpendicular to a chord, it divides the chord into two equal parts. Thus, Assertion (A) is True.

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Reason (R): This can be proven using congruent triangles formed by connecting the center to the endpoints of the chord and the foot of the perpendicular.

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Consider a circle with center O and a chord AB. Let M be the foot of the perpendicular from O to AB. We need to show that AM = MB. By connecting O to A and O to B, we form two right-angled triangles, $\triangle \text{O M A}$ and $\triangle \text{O M B}$ (since OM $\perp$ AB).

In $\triangle \text{OMA}$ and $\triangle \text{OMB}$:

• OA = OB (Radii of the same circle)
• OM = OM (Common side)
• $\angle \text{OMA} = \angle \text{OMB} = 90^\circ$ (Given that OM is perpendicular to AB)

Therefore, $\triangle \text{OMA} \cong \triangle \text{OMB}$ by the RHS (Right angle-Hypotenuse-Side) congruence rule.

By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), AM = MB.

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This demonstrates that the reason given (using congruent triangles) is indeed the standard and correct method to prove the assertion that the perpendicular from the center bisects the chord. Thus, Reason (R) is also True and is the correct explanation for Assertion (A).

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Since both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A), option (A) is correct.

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The correct option is (A) Both A and R are true and R is the correct explanation of A.

Assertion (A): A square is a cyclic quadrilateral.

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A cyclic quadrilateral is a quadrilateral whose all four vertices lie on the circumference of a circle. A property of cyclic quadrilaterals is that the sum of opposite angles is $180^\circ$. In a square, all four angles are $90^\circ$. The sum of any pair of opposite angles is $90^\circ + 90^\circ = 180^\circ$. Since this condition is satisfied, a square is indeed a cyclic quadrilateral. Thus, Assertion (A) is True.

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Reason (R): In a square, all angles are $90^\circ$, and the sum of opposite angles is $90^\circ + 90^\circ = 180^\circ$.

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This statement accurately describes the properties of a square concerning its angles. All interior angles of a square are right angles, i.e., $90^\circ$. Consequently, the sum of any two opposite angles is $90^\circ + 90^\circ = 180^\circ$. This property (sum of opposite angles being $180^\circ$) is precisely the condition that makes a quadrilateral cyclic. Therefore, Reason (R) is True and it correctly explains why Assertion (A) is true.

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Since both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A), option (A) is the correct choice.

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The correct option is (A) Both A and R are true and R is the correct explanation of A.

Let's match each term from Column A with its correct description in Column B:

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(i) Radius: The distance from the center of a circle to any point on its circumference. This definition matches description (c) "Distance from the center to the circumference". So, (i) corresponds to (c).

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(ii) Arc: A continuous portion of the circumference of a circle. This definition matches description (b) "A continuous piece of a circle". So, (ii) corresponds to (b).

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(iii) Sector: The region bounded by two radii of the circle and the intercepted arc between them. This definition matches description (a) "Region between two radii and an arc". So, (iii) corresponds to (a).

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(iv) Segment: The region bounded by a chord of the circle and the arc that the chord cuts off. This definition matches description (d) "Region between a chord and an arc". So, (iv) corresponds to (d).

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The correct matching is: (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d).

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Comparing this with the given options, we find that option (A) matches our result.

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The correct option is (A) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d).

Given:

A chord AB subtends an angle at the center O, $\angle \text{AOB} = 120^\circ$.

C is a point on the major arc AB.

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To Find:

The measure of the angle $\angle \text{ACB}$.

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Solution:

We use the theorem that states that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

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The minor arc AB subtends the angle $\angle \text{AOB}$ at the center and $\angle \text{ACB}$ at point C on the major arc (the remaining part of the circle).

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According to the theorem, the relationship between these angles is:

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Substitute the given value of $\angle \text{AOB}$:

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Now, solve for $\angle \text{ACB}$:

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Thus, the angle subtended by the chord AB at point C on the major arc is $60^\circ$.

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The correct option is (B) $60^\circ$.

This question is about the relationship between the length of a chord and its distance from the center of a circle.

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A theorem states that in a circle, the farther a chord is from the center, the shorter is its length. Conversely, the closer a chord is to the center, the longer is its length.

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In this case, we are given that chord AB is longer than chord CD ($AB > CD$).

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According to the theorem, a longer chord is closer to the center, and a shorter chord is farther from the center.

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Let the distance of chord AB from the center O be $d_{AB}$ and the distance of chord CD from the center O be $d_{CD}$.

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Since $AB > CD$, the chord AB is closer to the center than the chord CD.

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Therefore, the distance of AB from O is less than the distance of CD from O ($d_{AB} < d_{CD}$).

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The correct option is (B) Less than.

This question describes a fundamental theorem in circle geometry.

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The theorem states that the angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circle (i.e., on the circumference).

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Let $\theta_{\text{center}}$ be the angle at the center and $\theta_{\text{circumference}}$ be the angle at a point on the remaining part of the circle, both subtended by the same arc. The relationship is:

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Therefore, the angle subtended by an arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle.

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The correct option is (C) Double.

A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides.

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When the non-parallel sides of a trapezium are equal in length, the trapezium is given a special name.

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This specific type of trapezium is called an isosceles trapezium.

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Let's consider the other options:

(A) A parallelogram has both pairs of opposite sides parallel and equal in length, which is a different type of quadrilateral.

(B) A rectangle is a parallelogram with all four angles equal to $90^\circ$.

(D) An isosceles trapezium is always a cyclic quadrilateral, but the term "Isosceles trapezium" is the geometric classification based on the equality of its non-parallel sides.

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Therefore, if the non-parallel sides of a trapezium are equal, it is an isosceles trapezium.

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The correct option is (C) Isosceles trapezium.

A polygon is said to be inscribed in a circle if all its vertices lie on the circle.

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A polygon is called a cyclic polygon if there is a circle that passes through all its vertices. In other words, all the vertices of the polygon lie on the circumference of a circle.

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Let's examine the options:

(A) All its sides are equal describes an equilateral polygon. An equilateral polygon inscribed in a circle is cyclic, but the condition for being cyclic is not that all sides are equal (e.g., a rectangle is cyclic but its adjacent sides are not equal).

(B) All its angles are equal describes an equiangular polygon. An equiangular polygon inscribed in a circle is cyclic, but the condition for being cyclic is not that all angles are equal (e.g., an isosceles trapezium is cyclic, but its angles are not all equal).

(C) All its vertices lie on the circle is the precise definition of a cyclic polygon.

(D) All its diagonals pass through the center is generally not true for a polygon with more than 2 vertices. Only the diameter passes through the center. For a polygon, this would only happen for specific cases like a circle itself (which can be considered a degenerate polygon) or potentially some star polygons.

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Therefore, a polygon inscribed in a circle is called a cyclic polygon if all its vertices lie on the circle.

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The correct option is (C) All its vertices lie on the circle.

A quadrilateral is cyclic if and only if the sum of its opposite angles is $180^\circ$. We need to determine which of the given quadrilaterals always satisfies this property.

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(A) Parallelogram: In a parallelogram, opposite angles are equal. Let the angles be $\angle A, \angle B, \angle C, \angle D$. If it's a parallelogram, $\angle A = \angle C$ and $\angle B = \angle D$. The sum of all angles is $360^\circ$, so $\angle A + \angle B + \angle C + \angle D = 360^\circ$, which means $2\angle A + 2\angle B = 360^\circ$, or $\angle A + \angle B = 180^\circ$. For a parallelogram to be cyclic, the sum of opposite angles must be $180^\circ$. So, $\angle A + \angle C = 180^\circ$. Since $\angle A = \angle C$, this means $2\angle A = 180^\circ$, so $\angle A = 90^\circ$. Similarly, $2\angle B = 180^\circ$, so $\angle B = 90^\circ$. Thus, a parallelogram is cyclic if and only if all its angles are $90^\circ$, which means it must be a rectangle. Not all parallelograms are rectangles.

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(B) Rectangle: A rectangle is a parallelogram with all four angles equal to $90^\circ$. The opposite angles are $90^\circ$ and $90^\circ$. The sum of opposite angles is $90^\circ + 90^\circ = 180^\circ$. Since the sum of each pair of opposite angles is $180^\circ$, a rectangle is always a cyclic quadrilateral.

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(C) Rhombus: A rhombus is a parallelogram with all sides equal. Similar to a parallelogram, a rhombus is cyclic if and only if all its angles are $90^\circ$, which means it must be a square. Not all rhombuses are squares.

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(D) Trapezium: A trapezium (trapezoid) has at least one pair of parallel sides. The sum of opposite angles in a general trapezium is not necessarily $180^\circ$. Only an isosceles trapezium (where the non-parallel sides are equal) is always cyclic.

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Based on the analysis, a Rectangle is the only quadrilateral among the given options that is always cyclic.

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The correct option is (B) Rectangle.

Given:

A line segment joining two points, say A and B.

This line segment AB subtends equal angles at two other points, say C and D.

Points C and D are on the same side of the line segment AB.

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Concept:

This problem refers to the converse of the theorem regarding angles in the same segment of a circle.

The theorem states that angles in the same segment of a circle are equal.

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The converse of this theorem states that if a line segment joining two points (A and B) subtends equal angles ($\angle \text{ACB} = \angle \text{ADB}$) at two other points (C and D) lying on the same side of the line segment, then the four points A, B, C, and D lie on the same circle.

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Points that lie on the same circle are called concyclic points.

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Let's evaluate the options:

(A) Collinear means lying on the same straight line. This is not the case here.

(B) Concyclic means lying on the same circle. This matches the theorem's conclusion.

(C) Congruent describes figures with the same size and shape, not points.

(D) Parallel describes lines or segments that never intersect.

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Therefore, based on the converse of the theorem about angles in the same segment, if a line segment joining two points subtends equal angles at two other points on the same side of the line segment, then the four points are concyclic.

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The correct option is (B) Concyclic.

Given:

ABCD is a cyclic quadrilateral.

$\angle \text{A} = 70^\circ$.

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To Find:

The measure of $\angle \text{C}$.

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Solution:

A property of cyclic quadrilaterals states that the sum of a pair of opposite angles is $180^\circ$.

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In cyclic quadrilateral ABCD, the opposite angles are $\angle \text{A}$ and $\angle \text{C}$, and $\angle \text{B}$ and $\angle \text{D}$.

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Therefore, we have:

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Substitute the given value of $\angle \text{A}$ into the equation:

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Subtract $70^\circ$ from both sides to find $\angle \text{C}$:

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Thus, the measure of angle C is $110^\circ$.

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The correct option is (B) $110^\circ$.

Let's analyze each statement based on the definitions of the terms related to a circle:

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A chord is a line segment connecting any two points on the circumference of a circle.

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A diameter is a special type of chord that passes through the center of the circle. It is the longest chord in a circle.

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A radius is a line segment connecting the center of the circle to any point on the circumference.

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A tangent is a line that touches the circle at exactly one point.

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Now let's evaluate the given statements:

(A) A diameter is a chord: By definition, a diameter is a chord that goes through the center. Since it connects two points on the circle (the ends of the diameter), it satisfies the definition of a chord. This statement is TRUE.

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(B) A chord is a diameter: This is not always true. Only chords that pass through the center are diameters. Most chords do not pass through the center. This statement is FALSE.

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(C) A radius is a chord: A radius connects the center to a point on the circumference. A chord connects two points on the circumference. The center is not on the circumference (unless the radius is zero, which is a degenerate circle). So, a radius is not a chord. This statement is FALSE.

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(D) A chord is a tangent: A chord intersects the circle at two points, while a tangent intersects the circle at exactly one point. They are fundamentally different line segments/lines. This statement is FALSE.

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Therefore, the only true statement among the options is (A).

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The correct option is (A) A diameter is a chord.

When two distinct circles intersect at two points, let's call these points P and Q.

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A common chord of two circles is a line segment that is a chord of both circles.

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The line segment connecting the two points of intersection (P and Q) is a chord of the first circle because its endpoints lie on the first circle.

It is also a chord of the second circle because its endpoints lie on the second circle.

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Since there are exactly two points of intersection, there is only one unique line segment that connects these two points (the segment PQ).

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Therefore, there is only one common chord that can be drawn when two different circles intersect at two points.

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The correct option is (A) One.

A perpendicular bisector of a line segment is a line that is perpendicular to the segment and passes through its midpoint.

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Consider a chord AB of a circle with center O.

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The perpendicular bisector of the chord AB consists of all points that are equidistant from the endpoints A and B.

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The center O of the circle is equidistant from all points on the circumference. Since points A and B are on the circumference, the distance from the center O to A is equal to the distance from the center O to B (both are equal to the radius of the circle).

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Since the center O is equidistant from A and B, it must lie on the perpendicular bisector of the line segment AB.

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This is a fundamental theorem in circle geometry: The perpendicular bisector of any chord of a circle passes through the center of the circle.

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Therefore, the perpendicular bisector of a chord passes through the Center of the circle.

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The correct option is (C) Center.

Given:

ABCD is a cyclic quadrilateral.

$\angle \text{A} = 85^\circ$

$\angle \text{B} = 92^\circ$

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To Find:

The measures of $\angle \text{C}$ and $\angle \text{D}$.

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Solution:

In a cyclic quadrilateral, the sum of a pair of opposite angles is $180^\circ$.

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For the opposite angles $\angle \text{A}$ and $\angle \text{C}$, we have:

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Substitute the given value of $\angle \text{A}$ into equation (i):

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Solving for $\angle \text{C}$:

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For the opposite angles $\angle \text{B}$ and $\angle \text{D}$, we have:

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Substitute the given value of $\angle \text{B}$ into equation (ii):

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Solving for $\angle \text{D}$:

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Thus, the measures of angles C and D are $95^\circ$ and $88^\circ$, respectively.

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Comparing our results with the given options, we find that $\angle \text{C} = 95^\circ$ and $\angle \text{D} = 88^\circ$ matches option (A).

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The correct option is (A) $\angle C = 95^\circ, \angle D = 88^\circ$.

Let's examine each term provided in the options:

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(A) Vertex: A vertex is a point where two or more edges or lines meet, forming a corner. This term is commonly used in relation to polygons (like triangles, squares, etc.), polyhedra, angles, or graphs. Circles are smooth curves and do not have corners or vertices in this sense.

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(B) Chord: A chord is a line segment connecting two points on the circumference of a circle. This is a standard term used in circle geometry.

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(C) Arc: An arc is a continuous portion of the circumference of a circle. This is a fundamental term related to circles.

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(D) Segment: A segment of a circle is the region bounded by a chord and the arc subtended by the chord. This is also a standard term used in circle geometry.

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Based on these definitions, the term Vertex is not typically used in relation to circles, whereas Chord, Arc, and Segment are fundamental terms describing parts or regions of a circle.

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Therefore, Vertex is NOT a term related to circles.

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The correct option is (A) Vertex.

Given:

Distance of the chord from the center = 3 cm.

Radius of the circle = 5 cm.

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To Find:

The length of the chord.

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Solution:

Let the circle have center O and the chord be AB. Let M be the foot of the perpendicular drawn from the center O to the chord AB.

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According to the theorem, the perpendicular from the center of a circle to a chord bisects the chord. Therefore, M is the midpoint of AB, and AM = MB = $\frac{1}{2}$AB.

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We are given the distance of the chord from the center, which is OM = 3 cm.

The radius of the circle is OA = 5 cm (connecting the center to an endpoint of the chord).

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In the right-angled triangle $\triangle \text{OMA}$ (since OM $\perp$ AB):

By the Pythagorean theorem, we have:

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Substitute the given values:

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Subtract 9 from both sides:

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Taking the square root of both sides (and considering the positive value for length):

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Since M is the midpoint of AB, the length of the chord AB is 2 times AM:

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The length of the chord is 8 cm.

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The correct option is (C) 8 cm.

An arc (or the chord associated with it) divides a circle into two segments.

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Angles subtended by the same arc in the same segment of a circle have a specific relationship according to a theorem in circle geometry.

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The theorem states that angles in the same segment of a circle are equal.

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For example, if points C and D are on the same segment formed by chord AB (or arc AB), then the angle subtended by the arc AB at C ($\angle \text{ACB}$) is equal to the angle subtended by the arc AB at D ($\angle \text{ADB}$).

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Therefore, angles in the same segment of a circle are equal.

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The correct option is (C) Equal.

Let's review the definitions of the terms:

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(A) A Chord is a line segment connecting two points on the circumference.

(B) A Diameter is a chord that passes through the center.

(C) An Arc is a continuous portion of the circumference of a circle.

(D) A Segment is the region bounded by a chord and an arc.

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The question asks for the term that refers to "A part of the circumference of a circle".

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Based on the definitions, an Arc is precisely a part of the circumference.

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Therefore, a part of the circumference of a circle is called an Arc.

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The correct option is (C) Arc.

Given:

Two different circles intersect at two points.

The two circles are congruent.

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To Determine:

Properties of the common chord.

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Solution:

Let the two congruent circles be $C_1$ and $C_2$, with centers $O_1$ and $O_2$ respectively. Since the circles are congruent, they have the same radius, let's say $r$. Let the two points of intersection be A and B. The common chord is the line segment AB.

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Consider the line segment joining the centers $O_1O_2$. Connect the centers to the points of intersection.

In circle $C_1$, $O_1A$ and $O_1B$ are radii, so $O_1A = O_1B = r$.

In circle $C_2$, $O_2A$ and $O_2B$ are radii, so $O_2A = O_2B = r$.

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Thus, we have $O_1A = O_1B = O_2A = O_2B = r$.

The quadrilateral $O_1AO_2B$ has all four sides equal in length, which means $O_1AO_2B$ is a rhombus.

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The diagonals of the rhombus $O_1AO_2B$ are the line segment joining the centers $O_1O_2$ and the common chord AB.

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Properties of a rhombus include that its diagonals are perpendicular bisectors of each other.

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This means that the line segment $O_1O_2$ is perpendicular to the line segment AB, and $O_1O_2$ bisects AB. Also, the line segment AB is perpendicular to the line segment $O_1O_2$, and AB bisects $O_1O_2$.

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Let's check the given options:

(A) The common chord is a diameter in both circles. This would imply that the chord AB passes through both $O_1$ and $O_2$. Since the circles are different and intersect at two points, $O_1 \neq O_2$. A line segment cannot pass through two distinct points while also being a chord of two circles unless the circles are coaxial and the chord is the radical axis, but it wouldn't be a diameter of both circles generally. Also, the common chord is a diameter only if its length is $2r$. The distance between centers $O_1O_2$ for intersecting congruent circles is less than $2r$. If $M$ is the midpoint of $AB$, $O_1M \perp AB$. In $\triangle O_1MA$, $O_1A^2 = O_1M^2 + AM^2$. $r^2 = O_1M^2 + AM^2$. For AB to be a diameter, $AM=r$, which would imply $O_1M=0$, so the chord passes through the center. For it to be a common diameter, AB must pass through both centers, which is not possible for distinct circles intersecting at two points. So, this statement is false.

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(B) The line joining the centers is perpendicular to the common chord. As shown above, since $O_1AO_2B$ is a rhombus, its diagonals $O_1O_2$ and AB are perpendicular. This statement is True.

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(C) The common chord passes through the midpoint of the line joining the centers. As shown above, the diagonals of the rhombus $O_1AO_2B$ bisect each other. This means the intersection point of AB and $O_1O_2$ is the midpoint of both AB and $O_1O_2$. Therefore, the line segment AB passes through the midpoint of the line segment $O_1O_2$. This statement is True.

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(D) Both (B) and (C). Since both statement (B) and statement (C) are true for intersecting congruent circles, this combined statement is also true.

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The correct option is (D) Both (B) and (C).

A quadrilateral is called cyclic if all four of its vertices lie on the circumference of a circle. A key property of cyclic quadrilaterals is that the sum of each pair of opposite angles is $180^\circ$.

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Let's examine each type of quadrilateral:

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(A) Rectangle: A rectangle is a quadrilateral with all four angles equal to $90^\circ$. The opposite angles are $90^\circ$ and $90^\circ$. The sum of opposite angles is $90^\circ + 90^\circ = 180^\circ$. Since the sum of opposite angles is always $180^\circ$, a rectangle is always cyclic.

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(B) Rhombus: A rhombus is a parallelogram with all four sides equal. In a rhombus, opposite angles are equal, but they are not necessarily $90^\circ$. For a rhombus to be cyclic, its opposite angles must sum to $180^\circ$. If $\angle A = \angle C$ and $\angle B = \angle D$ are opposite angles, then $\angle A + \angle C = 180^\circ \implies 2\angle A = 180^\circ \implies \angle A = 90^\circ$. Similarly, $\angle B = 90^\circ$. Thus, a rhombus is cyclic if and only if all its angles are $90^\circ$, meaning it is a square. A general rhombus is not always cyclic.

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(C) Square: A square is a special type of rectangle and a special type of rhombus. It has all four sides equal and all four angles equal to $90^\circ$. As shown for a rectangle, the sum of opposite angles is $90^\circ + 90^\circ = 180^\circ$. Therefore, a square is always cyclic.

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(D) Isosceles Trapezium: An isosceles trapezium (or isosceles trapezoid) is a quadrilateral with one pair of parallel sides and the non-parallel sides are equal in length. A property of an isosceles trapezium is that its opposite angles are supplementary, meaning they add up to $180^\circ$. For instance, if AB is parallel to CD and AD = BC, then $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$. Since the sum of opposite angles is always $180^\circ$, an isosceles trapezium is always cyclic.

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Based on the analysis, Rectangles, Squares, and Isosceles Trapeziums are always cyclic quadrilaterals.

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The correct options are (A) Rectangle, (C) Square, and (D) Isosceles Trapezium.

Let the two points on the circumference of the circle be A and B.

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The distance between these two points can be measured along the curved path on the circumference or along a straight line connecting them.

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The curved path between A and B on the circumference is called an arc.

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The straight line segment connecting points A and B is called a chord.

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A fundamental principle of geometry states that the shortest distance between any two points in a plane is a straight line.

---

Therefore, the shortest distance between points A and B on the circumference is the length of the straight line segment connecting them, which is the chord AB.

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The other options are not general terms for the straight line segment between any two points on the circumference:

(C) A radius connects the center to a point on the circumference.

(D) A diameter is a specific chord that passes through the center.

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Thus, the shortest distance between two points on the circumference of a circle is along the Chord joining the points.

---

The correct option is (B) Chord.

Given:

Distance of the chord from the center = 4 cm.

Diameter of the circle = 10 cm.

---

To Find:

The length of the chord.

---

Solution:

First, find the radius of the circle. The radius (r) is half of the diameter.

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Let the circle have center O, and let AB be the chord. Let M be the foot of the perpendicular from O to AB.

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We are given the distance of the chord from the center, so OM = 4 cm.

The radius OA = 5 cm.

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In the right-angled triangle $\triangle \text{OMA}$, by the Pythagorean theorem:

---

Substitute the known values:

---

Subtract $16 \text{ cm}^2$ from both sides:

---

Taking the square root of both sides:

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The perpendicular from the center to a chord bisects the chord. So, AB = 2 $\times$ AM.

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The length of the chord is 6 cm.

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The correct option is (B) 6 cm.

An angle subtended by the diameter at any point on the circumference of a circle is a special case of the theorem about the angle subtended by an arc at the center and at the circumference.

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The diameter of a circle subtends a straight angle at the center, which measures $180^\circ$.

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The angle subtended by the same diameter at any point on the circumference (which is the remaining part of the circle) is half the angle subtended at the center.

---

According to the theorem:

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In this case, the angle at the center is $180^\circ$ (subtended by the diameter).

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This means the angle in a semicircle is always a right angle.

---

Therefore, the angle subtended by the diameter at any point on the circumference is $90^\circ$.

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The correct option is (B) $90^\circ$.

This question refers to a theorem relating angles at the center and the length of chords.

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The theorem states that if the angles subtended by two chords at the center of a circle (or of congruent circles) are equal, then the chords are equal in length.

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Conversely, if two chords of a circle (or of congruent circles) are equal, then the angles subtended by them at the center are equal.

---

Given that two chords of a circle subtend equal angles at the center, it directly follows from this theorem that the lengths of the chords must be equal.

---

Therefore, if two chords of a circle subtend equal angles at the center, then the chords are equal.

---

The correct option is (C) Equal.

Given:

ABCD is a cyclic quadrilateral on the circumference of a circle with center O.

The angle subtended by arc ABC at the center is $220^\circ$. This refers to the reflex angle $\angle \text{AOC}$.

---

To Find:

The measure of the angle $\angle \text{ADC}$.

---

Solution:

The angle $\angle \text{ADC}$ is the angle subtended by the arc ABC at point D on the remaining part of the circle.

---

We use the theorem which states that the angle subtended by an arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle.

---

In this case, the arc is ABC, which subtends the reflex angle $\angle \text{AOC}$ at the center and the angle $\angle \text{ADC}$ at point D on the remaining part of the circle.

---

According to the theorem:

---

Substitute the given value into equation (i):

---

Now, solve for $\angle \text{ADC}$:

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Alternatively, we can consider the angle subtended by the minor arc AC at the center and at point D. The angle subtended by the minor arc AC at the center is $\angle \text{AOC} = 360^\circ - \text{Reflex } \angle \text{AOC} = 360^\circ - 220^\circ = 140^\circ$. The angle subtended by the minor arc AC at point D is $\angle \text{ADC}$. This is not correct as D is on the major arc with respect to minor arc AC. Let's stick to the first approach as the arc ABC is clearly defined to subtend the reflex angle.

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The angle subtended by the arc ABC at point D is $\angle \text{ADC} = 110^\circ$.

---

The correct option is (B) $110^\circ$.

A circle is a collection of all the points in a plane which are at a fixed distance from a fixed point in the plane.

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The centre of the circle is the fixed point from which all the points on the circle are equidistant.

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The radius of the circle is the fixed distance between the centre and any point on the circle.

A chord of a circle is a line segment joining any two points on the circle.

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A diameter of a circle is a chord that passes through the centre of the circle.

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The diameter is the longest chord of a circle and its length is twice the length of the radius. Every diameter is a chord, but every chord is not a diameter (unless it passes through the centre).

An arc of a circle is a continuous piece of the circumference of the circle between two points on the circle.

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A sector of a circle is the region bounded by two radii and the arc intercepted between them. It is like a slice of pizza.

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To show a minor arc and a major arc, draw a circle with centre O. Mark two points P and Q on the circumference. The shorter part of the circumference between P and Q is the minor arc PQ. The longer part of the circumference between P and Q is the major arc PQ. A line segment PQ is the corresponding chord for both arcs.

A segment of a circle is the region bounded by a chord and the arc intercepted by the chord.

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A chord divides the circle into two regions, each called a segment. The segment bounded by the chord and the minor arc is called the minor segment.

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The segment bounded by the same chord and the major arc is called the major segment. The major segment is typically larger than the minor segment (unless the chord is a diameter, in which case both segments are semicircles and equal).

Two circles are said to be congruent if they can be made to coincide completely when one is placed on top of the other. In simpler terms, they are exact copies of each other.

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The congruence of circles is determined solely by their radii (or diameters). If two circles have the same radius (or diameter), they are congruent. Conversely, if two circles are congruent, they must have the same radius (or diameter).

Given:

A circle with centre O.

Two equal chords AB and CD.

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To Prove:

The angles subtended by the chords at the centre are equal, i.e., $\angle AOB = \angle COD$.

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Proof:

Consider the triangles $\triangle AOB$ and $\triangle COD$.

In $\triangle AOB$ and $\triangle COD$:

Therefore, by SSS congruence criterion,

Since the triangles are congruent, their corresponding parts are equal (CPCTC).

Thus, the angles subtended at the centre are equal:

Hence, equal chords of a circle subtend equal angles at the centre.

Given:

A circle with centre O.

Two chords AB and CD subtending equal angles at the centre, i.e., $\angle AOB = \angle COD$.

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To Prove:

The chords are equal, i.e., AB = CD.

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Proof:

Consider the triangles $\triangle AOB$ and $\triangle COD$.

In $\triangle AOB$ and $\triangle COD$:

Therefore, by SAS congruence criterion,

Since the triangles are congruent, their corresponding parts are equal (CPCTC).

Thus, the chords are equal:

Hence, if the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Given:

A circle with centre O.

Chord AB.

OM is perpendicular to AB, where M is a point on AB.

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To Prove:

The perpendicular OM bisects the chord AB, i.e., AM = MB.

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Construction Required:

Join OA and OB.

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Proof:

Consider the right-angled triangles $\triangle OMA$ and $\triangle OMB$.

(Since OM $\perp$ AB, $\angle OMA = \angle OMB = 90^\circ$)

In $\triangle OMA$ and $\triangle OMB$:

Therefore, by RHS (Right angle-Hypotenuse-Side) congruence criterion,

Since the triangles are congruent, their corresponding parts are equal (CPCTC).

Thus,

Hence, the perpendicular from the centre of a circle to a chord bisects the chord.

Given:

A circle with centre O.

Chord AB.

M is a point on AB such that M is the midpoint of AB, i.e., AM = MB.

OM is the line segment joining the centre O to the midpoint M.

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To Prove:

The line OM is perpendicular to the chord AB, i.e., $\angle OMA = \angle OMB = 90^\circ$.

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Construction Required:

Join OA and OB.

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Proof:

Consider the triangles $\triangle OMA$ and $\triangle OMB$.

In $\triangle OMA$ and $\triangle OMB$:

Therefore, by SSS congruence criterion,

Since the triangles are congruent, their corresponding parts are equal (CPCTC).

Thus,

Since AB is a straight line, $\angle OMA$ and $\angle OMB$ form a linear pair.

Substituting $\angle OMA = \angle OMB$, we get:

Since $\angle OMA = \angle OMB$, we also have $\angle OMB = 90^\circ$.

This means OM is perpendicular to AB.

Hence, the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Given:

A circle with centre O.

Two equal chords AB and CD, i.e., AB = CD.

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To Prove:

The chords AB and CD are equidistant from the centre O. This means the perpendicular distance from O to AB is equal to the perpendicular distance from O to CD.

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Construction Required:

Draw OM $\perp$ AB and ON $\perp$ CD. Join OA and OC.

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Proof:

We know that the perpendicular from the centre of a circle to a chord bisects the chord. Therefore, OM bisects AB and ON bisects CD.

Since AB = CD (Given), multiplying both sides by $\frac{1}{2}$ gives:

Now, consider the right-angled triangles $\triangle OMA$ and $\triangle ONC$.

(Since OM $\perp$ AB and ON $\perp$ CD, $\angle OMA = \angle ONC = 90^\circ$)

In $\triangle OMA$ and $\triangle ONC$:

Therefore, by RHS (Right angle-Hypotenuse-Side) congruence criterion,

Since the triangles are congruent, their corresponding parts are equal (CPCTC).

Thus,

OM and ON are the perpendicular distances of chords AB and CD from the centre O.

Hence, equal chords of a circle are equidistant from the centre.

Given:

A circle with centre O.

Two chords AB and CD.

The chords are equidistant from the centre, meaning the perpendicular distance from O to AB is equal to the perpendicular distance from O to CD. Let OM $\perp$ AB and ON $\perp$ CD, such that OM = ON.

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To Prove:

The chords are equal in length, i.e., AB = CD.

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Construction Required:

Join OA and OC.

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Proof:

We know that the perpendicular from the centre of a circle to a chord bisects the chord. Therefore, OM bisects AB and ON bisects CD.

This means M is the midpoint of AB, so AM = MB = $\frac{1}{2}$ AB.

And N is the midpoint of CD, so CN = ND = $\frac{1}{2}$ CD.

Now, consider the right-angled triangles $\triangle OMA$ and $\triangle ONC$.

(Since OM $\perp$ AB and ON $\perp$ CD, $\angle OMA = \angle ONC = 90^\circ$)

In $\triangle OMA$ and $\triangle ONC$:

Therefore, by RHS (Right angle-Hypotenuse-Side) congruence criterion,

Since the triangles are congruent, their corresponding parts are equal (CPCTC).

Thus,

We know that AM = $\frac{1}{2}$ AB and CN = $\frac{1}{2}$ CD. Substituting these into AM = CN:

Multiplying both sides by 2, we get:

Hence, chords equidistant from the centre of a circle are equal in length.

Given that AB and CD are two equal chords of a circle with centre O (AB = CD).

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According to the theorem which states that equal chords of a circle are equidistant from the centre, the distance of chord AB from the centre O will be equal to the distance of chord CD from the centre O.

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Let OM be the perpendicular distance from O to AB, and ON be the perpendicular distance from O to CD. Then, we can say that OM = ON.

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Justification:

The justification for this statement comes directly from the theorem mentioned above. We have already proven (in Question 10) that if two chords are equal in length, then their perpendicular distances from the centre are equal.

The point of intersection E does not affect the perpendicular distance of the chords from the centre. The distance of a chord from the centre is defined as the length of the perpendicular segment from the centre to the chord.

The theorem states that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

This means $\angle \text{at the centre} = 2 \times \angle \text{at the circumference}$ when both angles are subtended by the same arc.

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Part 1: Find $\angle PQR$ if $\angle PSR = 30^\circ$.

Both $\angle PQR$ and $\angle PSR$ are angles subtended by the same arc PR at points Q and S respectively, on the remaining part of the circle (specifically, in the same segment).

According to the property that angles subtended by the same arc in the same segment of a circle are equal:

$\angle PQR = \angle PSR$

Given that $\angle PSR = 30^\circ$.

Therefore, $\angle PQR = 30^\circ$.

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Regarding the note: If $\angle POR = 60^\circ$, find angle at circumference.

Here, $\angle POR$ is the angle at the centre subtended by arc PR, and the angle at the circumference subtended by the same arc is $\angle PQR$ (or $\angle PSR$).

Using the theorem: $\angle POR = 2 \times \angle PQR$.

Given $\angle POR = 60^\circ$.

$60^\circ = 2 \times \angle PQR$

$\angle PQR = \frac{60^\circ}{2}$

$\angle PQR = 30^\circ$

This confirms the relationship stated in the theorem and aligns with the result from the first part.

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Redo Part: Find $\angle POR$ if $\angle PQR = 50^\circ$.

Here, $\angle POR$ is the angle at the centre and $\angle PQR$ is the angle at the circumference, both subtended by the arc PR.

Using the theorem: $\angle POR = 2 \times \angle PQR$.

Given that $\angle PQR = 50^\circ$.

$\angle POR = 2 \times 50^\circ$

$\angle POR = 100^\circ$.

Given:

In a circle, chord AB subtends angles $\angle ACB$ and $\angle ADB$ at points C and D respectively, which lie in the same segment of the circle.

$\angle ACB = 40^\circ$

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To Find:

The measure of $\angle ADB$.

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Solution:

We are given a circle with chord AB.

Points C and D lie on the circumference in the same segment with respect to the chord AB.

Angles $\angle ACB$ and $\angle ADB$ are angles subtended by the same chord AB in the same segment of the circle.

According to the theorem, Angles in the same segment of a circle are equal.

Therefore,

We are given that $\angle ACB = 40^\circ$.

Substituting the given value, we get:

Thus, the measure of angle ADB is $40^\circ$.

The angle in a semicircle is a right angle, which measures $90^\circ$.

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Justification using the theorem from Question 13:

We assume the theorem from Question 13 is: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

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Given:

A semicircle. This means the arc considered is a semicircle, and the corresponding chord is a diameter of the circle.

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To Find:

The measure of the angle subtended by the semicircle at any point on the circumference, and to justify it using the given theorem.

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Solution:

Let AB be the diameter of the circle, with O being the center. The arc ACB is a semicircle.

The angle subtended by the arc ACB at the centre O is the straight angle $\angle AOB$.

Let C be any point on the remaining part of the circle (i.e., on the circumference). The angle subtended by the arc ACB at point C is $\angle ACB$. This is the angle in the semicircle.

According to the theorem from Question 13,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

So, for arc ACB and point C:

Substitute the value of $\angle AOB$:

Now, solve for $\angle ACB$:

Therefore, the angle in a semicircle is $90^\circ$ or a right angle.

Definition of a Cyclic Quadrilateral:

A quadrilateral is called a cyclic quadrilateral if all four of its vertices lie on the circumference of a circle.

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Sum of Opposite Angles of a Cyclic Quadrilateral:

The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$ (or two right angles).

If ABCD is a cyclic quadrilateral, then:

$\angle A + \angle C = 180^\circ$

$\angle B + \angle D = 180^\circ$

Given:

ABCD is a cyclic quadrilateral.

$\angle A = 70^\circ$

$\angle B = 110^\circ$

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To Find:

The measure of $\angle C$ and $\angle D$.

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Solution:

We know that in a cyclic quadrilateral, the sum of opposite angles is $180^\circ$.

For opposite angles $\angle A$ and $\angle C$:

Substitute the given value $\angle A = 70^\circ$:

Subtract $70^\circ$ from both sides:

For opposite angles $\angle B$ and $\angle D$:

Substitute the given value $\angle B = 110^\circ$:

Subtract $110^\circ$ from both sides:

Thus, the measure of $\angle C$ is $110^\circ$ and the measure of $\angle D$ is $70^\circ$.

If the sum of a pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic.

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Justification:

This is the converse of the property of a cyclic quadrilateral, which states that the sum of opposite angles of a cyclic quadrilateral is $180^\circ$.

Let ABCD be a quadrilateral.

Given:

$\angle A + \angle C = 180^\circ$ (or $\angle B + \angle D = 180^\circ$)

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Conclusion:

The quadrilateral ABCD is cyclic (i.e., its vertices A, B, C, and D lie on the circumference of a circle).

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Theorem:

If the sum of a pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic.

Given:

In a circle with centre O, AB is a chord.

Radius $OA = 5$ cm.

Distance of the chord AB from the centre O is 3 cm. Let M be the foot of the perpendicular from O to the chord AB. So, $OM = 3$ cm.

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To Find:

The length of the chord AB.

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Solution:

Since OM is the distance of the chord from the centre, OM is perpendicular to AB.

In the right-angled triangle $\triangle$OMA, OA is the hypotenuse, OM is the height, and AM is part of the chord.

Applying the Pythagorean theorem in $\triangle$OMA:

Substitute the given values $OA = 5$ cm and $OM = 3$ cm:

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Therefore, M is the midpoint of AB, and $AB = 2 \times AM$.

Substitute the value of AM:

The length of the chord AB is 8 cm.

Given:

In a circle with centre O, AB is a chord.

Radius $OA = 10$ cm.

Length of the chord $AB = 16$ cm.

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To Find:

The distance of the chord AB from the centre O.

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Solution:

Let M be the foot of the perpendicular from the centre O to the chord AB. The distance of the chord from the centre is OM.

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Therefore, M is the midpoint of AB, which means $AM = MB = \frac{1}{2} AB$.

Substitute the value of AB:

Now, consider the right-angled triangle $\triangle$OMA. OA is the radius (hypotenuse), AM is half the chord, and OM is the distance of the chord from the centre.

Applying the Pythagorean theorem in $\triangle$OMA:

Substitute the values $OA = 10$ cm and $AM = 8$ cm:

Rearrange the equation to find OM$^2$:

Take the square root of both sides:

The distance of the chord AB from the centre is 6 cm.

Two distinct circles can intersect at a maximum of two points.

The number of intersection points can be zero, one, or two.

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Here are the possible cases with descriptions of the diagrams:

Case 1: Zero Intersection Points

This occurs when the circles do not touch or overlap at all.

Diagram Description: Draw two separate circles. They can be either completely outside each other or one circle can be completely inside the other without touching.

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Case 2: One Intersection Point

This occurs when the two circles touch each other at exactly one point. This is also known as the circles being tangent to each other.

Diagram Description: Draw two circles touching externally at a single point. Or, draw two circles where one is inside the other, touching internally at a single point.

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Case 3: Two Intersection Points

This occurs when the two circles cross each other. The points of intersection are distinct.

Diagram Description: Draw two overlapping circles that cross each other. The points where the circumferences meet are the two intersection points.

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Two distinct circles cannot intersect at more than two points. If they intersected at three or more points, they would have to coincide and thus would not be distinct circles.

If two equal chords of a circle intersect inside the circle, then the segments of one chord are equal to the corresponding segments of the other chord.

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Given:

Let AB and CD be two equal chords of a circle with centre O, intersecting at point P inside the circle.

$\text{AB} = \text{CD}$

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To Conclude:

What can be said about the segments AP, PB, CP, and PD?

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Solution:

The segments of chord AB are AP and PB.

The segments of chord CD are CP and PD.

According to the theorem: If two equal chords of a circle intersect inside the circle, then the segments of one chord are equal to the corresponding segments of the other chord.

This means that:

The segment from one end of chord AB to the intersection point is equal to the segment from one end of chord CD to the intersection point, and similarly for the other ends.

Specifically, we can conclude that:

and

Alternatively, it could be $\text{AP} = \text{PD}$ and $\text{PB} = \text{CP}$, depending on how the points are labeled on the chords. However, the standard interpretation of the theorem states that the segments are equal pairwise, matching shortest with shortest and longest with longest.

Given:

In a circle with centre O, $\angle ABC = 40^\circ$, where A, B, and C are points on the circumference.

---

To Find:

The measure of $\angle AOC$, which is the angle subtended by arc AC at the centre.

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Solution:

The angle $\angle ABC$ is the angle subtended by the arc AC at a point B on the remaining part of the circle.

The angle $\angle AOC$ is the angle subtended by the same arc AC at the centre O.

According to the theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Using this theorem, the angle at the centre ($\angle AOC$) is twice the angle at the circumference ($\angle ABC$) subtended by the same arc AC.

Substitute the given value $\angle ABC = 40^\circ$:

Thus, the measure of $\angle AOC$ is $80^\circ$.

If a cyclic parallelogram is also a rhombus, it must be a square.

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Justification:

We are given a quadrilateral that is both a parallelogram and a cyclic quadrilateral.

Let the parallelogram be ABCD.

Since ABCD is a parallelogram:

Opposite angles are equal.

Since ABCD is a cyclic quadrilateral:

The sum of opposite angles is $180^\circ$.

Substitute $\angle C = \angle A$ into the equation $\angle A + \angle C = 180^\circ$:

Since $\angle A = \angle C$, $\angle C = 90^\circ$.

Similarly, substitute $\angle D = \angle B$ into the equation $\angle B + \angle D = 180^\circ$:

Since $\angle B = \angle D$, $\angle D = 90^\circ$.

Thus, a cyclic parallelogram must have all its interior angles equal to $90^\circ$. A parallelogram with all angles equal to $90^\circ$ is a rectangle.

Now, we are given that this figure is also a rhombus.

A rhombus is a parallelogram with all four sides equal in length.

So, the figure is a parallelogram with all angles $90^\circ$ (making it a rectangle) AND it has all sides equal (the property of a rhombus).

A quadrilateral that is both a rectangle (all angles $90^\circ$) and a rhombus (all sides equal) is by definition a square.

Therefore, if a cyclic parallelogram is a rhombus, it must be a square.

The statement "The shortest chord of a circle is the diameter" is false.

---

Justification:

A chord is a line segment connecting two points on the circumference of a circle.

A diameter is a special type of chord that passes through the centre of the circle.

Consider a circle with centre O and radius $r$. Let AB be any chord of the circle. Let M be the foot of the perpendicular from the centre O to the chord AB. The distance of the chord from the centre is $OM$.

In the right-angled triangle $\triangle$OMA, by the Pythagorean theorem:

$OA^2 = OM^2 + AM^2$

Since OA is the radius ($r$), we have:

$r^2 = OM^2 + AM^2$

$AM^2 = r^2 - OM^2$

$AM = \sqrt{r^2 - OM^2}$

The perpendicular from the centre bisects the chord, so $AB = 2 \times AM$.

$AB = 2\sqrt{r^2 - OM^2}$

From this formula, we can see how the length of the chord AB changes with the distance OM from the centre:

• If OM increases, $OM^2$ increases, so $r^2 - OM^2$ decreases, and $\sqrt{r^2 - OM^2}$ decreases. Therefore, the length of the chord AB decreases.
• If OM decreases, $OM^2$ decreases, so $r^2 - OM^2$ increases, and $\sqrt{r^2 - OM^2}$ increases. Therefore, the length of the chord AB increases.

The distance OM can range from 0 (when the chord passes through the centre) up to almost $r$ (when the chord becomes very short, approaching a point on the circumference).

The chord with the minimum distance from the centre is the diameter, where $OM = 0$. For the diameter, the length is $2\sqrt{r^2 - 0^2} = 2\sqrt{r^2} = 2r$, which is the maximum possible length of a chord.

The shortest chords are those that are furthest from the centre (i.e., have the largest possible distance OM, close to $r$). As the distance OM approaches $r$, the term $r^2 - OM^2$ approaches 0, and the length of the chord $2\sqrt{r^2 - OM^2}$ approaches 0.

Therefore, the diameter is the longest chord of a circle, not the shortest.

Theorem: Equal chords of a circle subtend equal angles at the centre.

---

Given:

A circle with centre O.

Two chords AB and CD such that $AB = CD$.

---

To Prove:

The angles subtended by these chords at the centre are equal, i.e., $\angle AOB = \angle COD$.

---

Proof:

Consider triangles $\triangle AOB$ and $\triangle COD$.

Therefore, by SSS congruence rule,

Since the triangles are congruent, their corresponding parts are equal (CPCT).

Hence, equal chords of a circle subtend equal angles at the centre.

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Converse of the Theorem:

The converse of the theorem is: If the angles subtended by two chords at the centre of a circle are equal, then the chords are equal.

---

Given:

A circle with centre O.

Two chords AB and CD such that the angles subtended by them at the centre are equal, i.e., $\angle AOB = \angle COD$.

---

To Prove:

The chords are equal, i.e., $AB = CD$.

---

Proof:

Consider triangles $\triangle AOB$ and $\triangle COD$.

Therefore, by SAS congruence rule,

Since the triangles are congruent, their corresponding parts are equal (CPCT).

Hence, if the angles subtended by two chords at the centre of a circle are equal, then the chords are equal.

Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.

---

Given:

A circle with centre O.

A chord AB of the circle.

OM is drawn perpendicular from O to chord AB, where M is a point on AB.

---

To Prove:

The chord AB is bisected by OM, i.e., $AM = MB$.

---

Construction Required:

Join OA and OB. (OA and OB are the radii of the circle).

Diagram Description: Draw a circle with centre O. Draw a chord AB. Draw a line segment OM from O perpendicular to AB, meeting AB at M. Draw line segments OA and OB connecting the centre to the endpoints of the chord.

---

Proof:

Consider the right-angled triangles $\triangle OMA$ and $\triangle OMB$.

In $\triangle OMA$, $\angle OMA = 90^\circ$ (since OM $\perp$ AB).

In $\triangle OMB$, $\angle OMB = 90^\circ$ (since OM $\perp$ AB).

So, we have:

Therefore, by RHS (Right angle - Hypotenuse - Side) congruence rule,

Since the triangles are congruent, their corresponding parts are equal (CPCT).

This shows that M is the midpoint of AB.

Hence, the perpendicular from the centre of a circle to a chord bisects the chord.

Theorem: Equal chords of a circle are equidistant from the centre.

---

Given:

A circle with centre O.

Two equal chords AB and CD, i.e., $AB = CD$.

Let OM $\perp$ AB and ON $\perp$ CD, where M is a point on AB and N is a point on CD.

---

To Prove:

The chords are equidistant from the centre, i.e., $OM = ON$.

---

Construction Required:

Join OA and OC. (OA and OC are the radii of the circle).

Diagram Description: Draw a circle with centre O. Draw two equal chords AB and CD. Draw perpendiculars OM from O to AB and ON from O to CD. Join OA and OC.

---

Proof:

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Since OM $\perp$ AB, M is the midpoint of AB.

Since ON $\perp$ CD, N is the midpoint of CD.

Given that $AB = CD$, dividing both sides by 2, we get $\frac{1}{2} AB = \frac{1}{2} CD$.

Now consider the right-angled triangles $\triangle OMA$ and $\triangle ONC$.

In $\triangle OMA$, $\angle OMA = 90^\circ$ (by construction, as OM is the distance).

In $\triangle ONC$, $\angle ONC = 90^\circ$ (by construction, as ON is the distance).

So, we have:

Therefore, by RHS (Right angle - Hypotenuse - Side) congruence rule,

Since the triangles are congruent, their corresponding parts are equal (CPCT).

Hence, equal chords of a circle are equidistant from the centre.

---

Converse of the Theorem: If the chords are equidistant from the centre, then the chords are equal.

---

Given:

A circle with centre O.

Two chords AB and CD.

Let OM $\perp$ AB and ON $\perp$ CD, where M is on AB and N is on CD.

The chords are equidistant from the centre, i.e., $OM = ON$.

---

To Prove:

The chords are equal, i.e., $AB = CD$.

---

Construction Required:

Join OA and OC. (OA and OC are the radii of the circle).

Diagram Description: Draw a circle with centre O. Draw two chords AB and CD. Draw perpendiculars OM from O to AB and ON from O to CD, such that OM = ON. Join OA and OC.

---

Proof:

Consider the right-angled triangles $\triangle OMA$ and $\triangle ONC$.

In $\triangle OMA$, $\angle OMA = 90^\circ$ (by construction, as OM is the distance).

In $\triangle ONC$, $\angle ONC = 90^\circ$ (by construction, as ON is the distance).

So, we have:

Therefore, by RHS (Right angle - Hypotenuse - Side) congruence rule,

Since the triangles are congruent, their corresponding parts are equal (CPCT).

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Since OM $\perp$ AB, $AB = 2 \times AM$.

Since ON $\perp$ CD, $CD = 2 \times CN$.

Since $AM = CN$, multiplying both sides by 2, we get $2 \times AM = 2 \times CN$.

Hence, if the chords are equidistant from the centre, then the chords are equal.

Theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

---

Given:

A circle with centre O.

An arc PQ of the circle.

R is a point on the remaining part of the circle.

---

To Prove:

$\angle POQ = 2 \angle PRQ$, where $\angle POQ$ is the angle subtended by arc PQ at the centre and $\angle PRQ$ is the angle subtended by arc PQ at point R on the remaining part of the circle.

---

Construction Required:

Join the point R to the centre O and extend RO to a point S such that S lies outside the circle.

Diagram Description: Draw a circle with centre O. Draw an arc PQ. Mark a point R on the circumference different from P and Q. Join OP, OQ, and OR. Extend RO to a point S outside the circle.

---

Proof:

We will consider three cases based on the position of point R and the nature of the arc PQ.

Case 1: The arc PQ is a minor arc and R is on the major arc.

In $\triangle$ROP, OA = OP (Radii of the same circle). Therefore, $\triangle$ROP is an isosceles triangle.

The exterior angle $\angle POS$ of $\triangle$ROP is equal to the sum of the two interior opposite angles.

Substituting $\angle OPR = \angle ORP$:

Similarly, in $\triangle$ROQ, OA = OQ (Radii of the same circle). Therefore, $\triangle$ROQ is an isosceles triangle.

The exterior angle $\angle SOQ$ of $\triangle$ROQ is equal to the sum of the two interior opposite angles.

Substituting $\angle OQR = \angle ORQ$:

Now, add the equations for $\angle POS$ and $\angle SOQ$:

The sum $\angle POS + \angle SOQ$ is equal to $\angle POQ$.

The sum $\angle ORP + \angle ORQ$ is equal to $\angle PRQ$.

Thus, the theorem is proved for Case 1.

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Case 2: The arc PQ is a semicircle.

In this case, PQ is a diameter of the circle, and the centre O lies on the chord PQ.

The angle subtended by the semicircle at the centre is the straight angle $\angle POQ$.

Let R be any point on the circumference. $\angle PRQ$ is the angle in the semicircle.

From Case 1, the relationship $\angle POQ = 2 \angle PRQ$ holds.

Substituting $\angle POQ = 180^\circ$:

This shows that the angle in a semicircle is a right angle, and the relationship $\angle POQ = 2 \angle PRQ$ holds as $180^\circ = 2 \times 90^\circ$.

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Case 3: The arc PQ is a major arc and R is on the minor arc.

In this case, the angle subtended by the major arc PQ at the centre is the reflex angle $\angle POQ$.

Let R be a point on the minor arc. The angle subtended by the major arc PQ at R is $\angle PRQ$.

Using the same construction (join RO and extend to S):

In $\triangle$ROP, $\angle POS = 2 \angle ORP$ (as shown in Case 1).

In $\triangle$ROQ, $\angle SOQ = 2 \angle ORQ$ (as shown in Case 1).

The reflex angle $\angle POQ$ is the sum $\angle POS + \angle SOQ$.

The sum $\angle ORP + \angle ORQ$ is equal to $\angle PRQ$.

Thus, the theorem is proved for Case 3.

In all cases, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Theorem: Angles in the same segment of a circle are equal.

---

Given:

A circle with centre O.

An arc PQ.

Two points R and S on the remaining part of the circle (i.e., in the same segment).

The angles subtended by the arc PQ at R and S are $\angle PRQ$ and $\angle PSQ$ respectively.

---

To Prove:

$\angle PRQ = \angle PSQ$.

---

Construction Required:

Join OP and OQ. (OP and OQ are radii).

Diagram Description: Draw a circle with centre O. Draw an arc PQ. Mark two points R and S on the circumference such that they are in the same segment with respect to arc PQ. Join PR, QR, PS, and QS. Join OP and OQ.

---

Proof:

The angle subtended by the arc PQ at the centre is $\angle POQ$.

According to the theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Using this theorem, the angle subtended by arc PQ at point R on the remaining part of the circle is $\angle PRQ$.

This can be rewritten as:

Similarly, the angle subtended by the same arc PQ at point S on the remaining part of the circle is $\angle PSQ$.

This can also be rewritten as:

From equations (i) and (ii), we see that both $\angle PRQ$ and $\angle PSQ$ are equal to half of the same angle $\angle POQ$.

Therefore,

Hence, angles in the same segment of a circle are equal.

Theorem: The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$.

---

Given:

ABCD is a cyclic quadrilateral, i.e., its vertices A, B, C, and D lie on the circumference of a circle with centre O.

---

To Prove:

$\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$.

---

Construction Required:

Join OB and OD.

Diagram Description: Draw a circle with centre O. Draw a cyclic quadrilateral ABCD. Join OB and OD. This divides the circle into two arcs BCD and BAD.

---

Proof:

Consider the arc BCD. It subtends $\angle BOD$ at the centre and $\angle BAD$ (which is $\angle A$) at the point A on the remaining part of the circle.

According to the theorem: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

So,

(Here, $\angle BOD$ is the angle subtended by the minor arc BCD at the centre).

Now consider the arc BAD. It subtends the reflex angle $\angle BOD$ at the centre and $\angle BCD$ (which is $\angle C$) at the point C on the remaining part of the circle.

Using the same theorem:

The sum of the angle and the reflex angle around the centre O is $360^\circ$.

Substitute the values from (i) and (ii):

Divide both sides by 2:

Similarly, we can prove that $\angle B + \angle D = 180^\circ$ by considering arcs ADC and ABC and joining OA and OC.

Arc ADC subtends $\angle AOC$ at the centre and $\angle ABC$ (which is $\angle B$) at point B on the remaining part.

Arc ABC subtends the reflex angle $\angle AOC$ at the centre and $\angle ADC$ (which is $\angle D$) at point D on the remaining part.

Adding these two equations:

Hence, the sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$.

Theorem: If the sum of a pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic.

---

Given:

A quadrilateral ABCD such that the sum of one pair of opposite angles is $180^\circ$.

(Note: If one pair of opposite angles sums to $180^\circ$, the other pair must also sum to $180^\circ$ since the sum of all angles in a quadrilateral is $360^\circ$).

---

To Prove:

The quadrilateral ABCD is cyclic, i.e., its vertices A, B, C, and D lie on the circumference of a circle.

---

Construction Required:

Draw a circle passing through three non-collinear points A, B, and C. Let this circle be $\mathcal{C}$.

Diagram Description: Draw a quadrilateral ABCD. Draw a circle passing through vertices A, B, and C. Let this circle be $\mathcal{C}$.

---

Proof:

Let the circle passing through points A, B, and C be denoted by $\mathcal{C}$.

Consider the chord AC. The angle subtended by the arc ABC at the circumference is $\angle ADC$. The angle subtended by the arc ADC at the circumference is $\angle ABC$.

We are given that in quadrilateral ABCD:

Now, let D' be any point on the circle $\mathcal{C}$ such that D' lies on the same side of the line segment AC as vertex D.

Since the points A, B, C, and D' all lie on the circle $\mathcal{C}$, the quadrilateral ABCD' is a cyclic quadrilateral.

According to the theorem about cyclic quadrilaterals (Question 6), the sum of opposite angles is $180^\circ$.

Therefore, for the cyclic quadrilateral ABCD':

Comparing the given condition ($\angle ABC + \angle ADC = 180^\circ$) and the property of the cyclic quadrilateral ABCD' ($\angle ABC + \angle AD'C = 180^\circ$), we can equate the second terms:

Now we have two points D and D' that lie on the same side of the line segment AC, and they subtend equal angles at the endpoints of the segment AC ($\angle ADC = \angle AD'C$).

A known geometric property states that if a line segment joining two points subtends equal angles at two other points lying on the same side of the line segment, then the four points are concyclic (lie on the same circle).

Applying this property to the line segment AC and points D and D': Since $\angle ADC = \angle AD'C$ and D and D' are on the same side of AC, the points A, C, D, and D' are concyclic.

By construction, the points A, C, and D' already lie on the circle $\mathcal{C}$. Since A, C, D, and D' are concyclic, the point D must also lie on the same circle $\mathcal{C}$ that passes through A, C, and D'.

Thus, all four vertices A, B, C, and D lie on the circle $\mathcal{C}$.

Therefore, ABCD is a cyclic quadrilateral.

Given:

A circle with centre O and radius $r = 5$ cm.

AB and AC are two chords.

Length of chord $AB = 6$ cm.

Length of chord $AC = 8$ cm.

---

To Find:

The distance of the chord BC from the centre O.

---

Solution:

The radius of the circle is $r = 5$ cm. The diameter of the circle is $d = 2 \times r = 2 \times 5 = 10$ cm.

Consider the triangle ABC formed by the two chords AB, AC, and the chord BC.

Let's examine the lengths of the chords AB and AC and compare the sum of the squares of their lengths with the square of the diameter.

Sum of the squares of the lengths of chords AB and AC:

Now consider the square of the diameter:

We observe that $\text{AB}^2 + \text{AC}^2 = 100$. If BC were the third side of the triangle, and $\angle BAC$ was a right angle, then by the Pythagorean theorem, $\text{BC}^2 = \text{AB}^2 + \text{AC}^2 = 100$. This would mean $BC = \sqrt{100} = 10$ cm.

If the length of chord BC is 10 cm, it is equal to the diameter of the circle.

We know that the angle subtended by a diameter at any point on the circumference is $90^\circ$. If A is a point on the circumference, and BC is a chord such that $\angle BAC = 90^\circ$, then BC must be the diameter.

Conversely, if $\text{AB}^2 + \text{AC}^2 = \text{BC}^2$, then $\triangle ABC$ is a right-angled triangle with the right angle at A. In this case, BC is the hypotenuse.

If BC has a length of 10 cm, which is the diameter, then it must pass through the centre O.

Therefore, BC is the diameter of the circle.

The distance of the diameter from the centre of the circle is always zero.

Thus, the distance of the chord BC from the centre is 0 cm.

Given:

Two circles intersect at points A and B. AD and AC are diameters of the two circles.

---

To Prove:

The points D, B, and C are collinear.

---

Proof:

Consider the circle for which AD is the diameter.

Since B is a point on the circumference of this circle, the angle subtended by the diameter AD at point B is a right angle.

---

Now, consider the circle for which AC is the diameter.

Since B is also a point on the circumference of this circle, the angle subtended by the diameter AC at point B is a right angle.

---

Consider the sum of angles $\angle ABD$ and $\angle ABC$. Both angles share the common vertex B and the common arm AB.

---

Since the sum of the adjacent angles $\angle ABD$ and $\angle ABC$ is $180^\circ$, they form a linear pair. This means that the rays BD and BC lie on a straight line passing through B.

Therefore, the points D, B, and C are collinear.

Hence Proved.

Given:

O is the centre of the circle.

$\angle OAB = 30^\circ$

$\angle OCB = 40^\circ$

---

To Find:

$\angle AOC$ and $\angle ABC$

---

Solution:

In $\triangle OAB$, we have OA = OB (Radii of the same circle).

Since OA = OB, $\triangle OAB$ is an isosceles triangle.

Therefore, the angles opposite to equal sides are equal.

---

In $\triangle OCB$, we have OB = OC (Radii of the same circle).

Since OB = OC, $\triangle OCB$ is an isosceles triangle.

Therefore, the angles opposite to equal sides are equal.

---

Now, we can find $\angle ABC$ by adding $\angle OBA$ and $\angle OBC$.

---

In $\triangle OAB$, the sum of angles is $180^\circ$.

---

In $\triangle OCB$, the sum of angles is $180^\circ$.

---

Now, we can find $\angle AOC$ by adding $\angle AOB$ and $\angle BOC$.

---

Alternatively, we know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

The reflex $\angle AOC$ subtends the major arc AC at the centre.

The angle $\angle ABC$ subtends the major arc AC at point B on the remaining part of the circle.

The angle $\angle AOC$ (non-reflex) is $360^\circ$ minus the reflex angle.

Both methods give the same result for $\angle AOC$.

---

Final Answer:

$\angle AOC = 220^\circ$

$\angle ABC = 70^\circ$

Given:

A trapezium ABCD such that AB || DC and the non-parallel sides are equal, i.e., AD = BC.

---

To Prove:

Trapezium ABCD is cyclic.

---

Construction Required:

Draw AE $\perp$ DC and BF $\perp$ DC, where E and F lie on DC.

---

Proof:

Since AB || DC and AE $\perp$ DC, BF $\perp$ DC, it follows that AB || EF and AE || BF.

Also, $\angle AEF = 90^\circ$ and $\angle BFE = 90^\circ$.

Thus, ABFE is a rectangle.

In rectangle ABFE, opposite sides are equal.

---

Now consider the right-angled triangles $\triangle ADE$ and $\triangle BCF$.

We have:

By the RHS congruence criterion, $\triangle ADE \cong \triangle BCF$.

Therefore, the corresponding parts are equal.

---

Since AB || DC, the sum of consecutive interior angles is $180^\circ$.

---

We know that $\angle D = \angle C$. Substitute $\angle C$ for $\angle D$ in the first equation:

Similarly, substitute $\angle D$ for $\angle C$ in the second equation:

---

We have shown that the sum of opposite angles of the trapezium ABCD is $180^\circ$ ($\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$).

A quadrilateral is cyclic if and only if the sum of opposite angles is $180^\circ$.

Therefore, trapezium ABCD is cyclic.

Hence Proved.

Let's consider a circle with centre O and an arc AB.

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Angle Subtended by an Arc at the Centre:

This is the angle formed by joining the endpoints of the arc (A and B) to the centre of the circle (O). It is the angle $\angle AOB$. This angle is unique for a given arc in a specific circle.

---

Angle Subtended by the Same Arc at a Point on the Circumference:

This is the angle formed by joining the endpoints of the same arc (A and B) to any point (say, P) on the remaining part of the circumference of the circle. It is the angle $\angle APB$. The "remaining part of the circumference" means the part of the circle that does not contain the arc AB itself.

---

Relationship between the Angles:

There is a fundamental theorem relating these two angles:

The angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the remaining part of the circle.

Mathematically, for arc AB and a point P on the remaining part of the circle:

$\angle AOB = 2 \times \angle APB$

---

Effect of the Position of the Point on the Circumference:

The position of the point P on the circumference does not affect the measure of the angle $\angle APB$, as long as P is on the remaining part of the circle subtended by the arc AB.

If you take any other point Q on the same remaining part of the circle, the angle $\angle AQB$ subtended by the arc AB at Q will be equal to $\angle APB$.

This is a direct consequence of the theorem $\angle AOB = 2 \times \angle APB$. Since $\angle AOB$ is fixed for a given arc, $\angle APB = \frac{1}{2} \angle AOB$ must also be fixed for any point P on the remaining arc.

However, if the point is on the *minor* arc when the major arc subtends the angle at the center, the angle will be supplementary to the angle subtended on the major arc. But conventionally, the theorem refers to the angle subtended on the *remaining* part, which is consistent.

---

In summary, the angle at the centre is twice the angle at the circumference (on the remaining arc), and the angle at the circumference is constant for all points on that remaining arc.