Circles: Cyclic Quadrilaterals

Cyclic Quadrilateral: Definition

In the study of quadrilaterals, a special category arises when all four corners, or vertices, of the quadrilateral lie precisely on the circumference of a single circle. Such a quadrilateral is known as a cyclic quadrilateral.

Formally, a quadrilateral ABCD is termed cyclic if and only if there exists a circle that passes through all four points A, B, C, and D simultaneously. These points are then referred to as concyclic points.

The unique circle that contains all four vertices of a cyclic quadrilateral is called its circumcircle or circumscribed circle. The centre of this circle is known as the circumcentre of the cyclic quadrilateral.

It is important to note that not every quadrilateral is cyclic. For instance, a general parallelogram is not cyclic unless it possesses $90^\circ$ angles, making it a rectangle or a square (both of which are always cyclic). Similarly, a kite is not cyclic unless the sum of a pair of opposite angles is $180^\circ$, which happens only under specific geometric conditions (e.g., if it has two right angles at opposite vertices where diagonals intersect).

Being cyclic imposes strong constraints on the properties of a quadrilateral, particularly regarding its angles, as we will see in the following sections.

Property of Opposite Angles of a Cyclic Quadrilateral (Theorem and Converse)

One of the most fundamental and useful properties of cyclic quadrilaterals is the relationship between their opposite angles.

Theorem: The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$ (i.e., they are supplementary).

This means that if a quadrilateral ABCD is cyclic, then $\angle \text{A} + \angle \text{C} = 180^\circ$ and $\angle \text{B} + \angle \text{D} = 180^\circ$.

Given: A cyclic quadrilateral ABCD inscribed in a circle with centre O.

To Prove: $\angle \text{A} + \angle \text{C} = 180^\circ$ and $\angle \text{B} + \angle \text{D} = 180^\circ$.

Construction: Join the centre O to vertices B and D. This creates angles $\angle \text{BOD}$ and Reflex $\angle \text{BOD}$ at the centre.

Proof:

We use the theorem that 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.

Consider the arc BCD. This arc subtends Reflex $\angle \text{BOD}$ at the centre O and $\angle \text{BAD}$ (or $\angle \text{A}$) at the point A on the remaining part of the circle.

Now consider the arc DAB (the minor arc BD). This arc subtends $\angle \text{BOD}$ at the centre O and $\angle \text{BCD}$ (or $\angle \text{C}$) at the point C on the remaining part of the circle.

Adding equation (i) and equation (ii):

Reflex $\angle \text{BOD} + \angle \text{BOD} = 2 \angle \text{A} + 2 \angle \text{C}$

The sum of an angle and its corresponding reflex angle at a point is always $360^\circ$. Therefore, Reflex $\angle \text{BOD} + \angle \text{BOD} = 360^\circ$.

Substituting this into the equation:

$360^\circ = 2 (\angle \text{A} + \angle \text{C})$

Dividing both sides by 2:

$\angle \text{A} + \angle \text{C} = \frac{360^\circ}{2} = 180^\circ$

Since the sum of interior angles in any quadrilateral is $360^\circ$, we have $\angle \text{A} + \angle \text{B} + \angle \text{C} + \angle \text{D} = 360^\circ$.

We can group the terms:

$(\angle \text{A} + \angle \text{C}) + (\angle \text{B} + \angle \text{D}) = 360^\circ$

Substituting the value we found for $(\angle \text{A} + \angle \text{C})$:

$180^\circ + (\angle \text{B} + \angle \text{D}) = 360^\circ$

Subtracting $180^\circ$ from both sides:

$\angle \text{B} + \angle \text{D} = 360^\circ - 180^\circ = 180^\circ$

Thus, the sum of both pairs of opposite angles of a cyclic quadrilateral is $180^\circ$.

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

This means that if in a quadrilateral ABCD, $\angle \text{A} + \angle \text{C} = 180^\circ$ (or $\angle \text{B} + \angle \text{D} = 180^\circ$), then the vertices A, B, C, and D must lie on a single circle.

Given: A quadrilateral ABCD such that $\angle \text{A} + \angle \text{C} = 180^\circ$.

To Prove: Quadrilateral ABCD is cyclic (i.e., points A, B, C, D are concyclic).

Proof: We will prove this by the method of contradiction.

We know that any three non-collinear points define a unique circle. Let's consider the points A, B, and D, which are vertices of the quadrilateral and are non-collinear (otherwise ABCD would degenerate into a triangle). We can draw a unique circle passing through A, B, and D.

Now, assume for the sake of contradiction that the fourth point C does not lie on this circle passing through A, B, and D.

If C is not on the circle, it must be either in the interior or the exterior of the circle.

Case 1: Point C lies in the exterior of the circle passing through A, B, and D.

Let the line segment DC intersect the circle at a point C' (Since D is on the circle and C is outside, the line segment DC must intersect the circle at some point C' between D and C).

Now, the quadrilateral ABC'D is cyclic because all its vertices (A, B, C', D) lie on the circle.

By the cyclic quadrilateral theorem we just proved, the sum of opposite angles of ABC'D is $180^\circ$.

We are given that in quadrilateral ABCD, the sum of opposite angles $\angle \text{A}$ and $\angle \text{C}$ is $180^\circ$.

Comparing equation (v) and equation (vi), we get:

$\angle \text{BC'D} = \angle \text{BCD}$

Now, consider $\triangle \text{BC'C}$. The angle $\angle \text{BCD}$ is an exterior angle to $\triangle \text{BC'C}$ at vertex C' (since C' lies on the segment DC). No, this is incorrect based on the diagram and assumption that C' is between D and C. Let's use the exterior angle theorem correctly.

In $\triangle \text{BC'C}$, $\angle \text{BCD}$ is the interior angle at C. The angle $\angle \text{BC'D}$ is on the straight line segment DC. The angles $\angle \text{BC'D}$ and $\angle \text{BC'C}$ are supplementary because they form a linear pair:

$\angle \text{BC'D} + \angle \text{BC'C} = 180^\circ$

In $\triangle \text{BC'C}$, the sum of interior angles is $180^\circ$:

$\angle \text{C'BC} + \angle \text{BC'C} + \angle \text{BCC'} = 180^\circ$

Note that $\angle \text{BCC'} = \angle \text{BCD}$. So, $\angle \text{C'BC} + \angle \text{BC'C} + \angle \text{BCD} = 180^\circ$

We had $\angle \text{BC'D} = \angle \text{BCD}$. Substitute $\angle \text{BCD}$ with $\angle \text{BC'D}$ in the triangle sum equation:

$\angle \text{C'BC} + \angle \text{BC'C} + \angle \text{BC'D} = 180^\circ$

Group the last two terms:

$\angle \text{C'BC} + (\angle \text{BC'C} + \angle \text{BC'D}) = 180^\circ$

Since $\angle \text{BC'C} + \angle \text{BC'D} = 180^\circ$ (linear pair), we substitute this:

$\angle \text{C'BC} + 180^\circ = 180^\circ$

This implies $\angle \text{C'BC} = 0^\circ$. An angle of $0^\circ$ in a triangle means the points C', B, and C are collinear. However, C' and B are distinct points on the circle through A, B, D, and C is a point on the line DC outside the circle. If C, B, C' were collinear, then B must lie on the line DC. But B is a vertex of a quadrilateral ABCD, not a point on the line DC (unless it's a degenerate case). This contradicts the fact that ABCD is a quadrilateral. Therefore, our assumption that C lies outside the circle must be false.

Case 2: Point C lies in the interior of the circle passing through A, B, and D.

Extend the line segment DC to intersect the circle at a point C' (Since D is on the circle and C is inside, the line segment DC extended must intersect the circle at some point C' such that C is between D and C').

Again, ABC'D is a cyclic quadrilateral as its vertices A, B, C', and D lie on the circle.

By the cyclic quadrilateral theorem:

We are given that in quadrilateral ABCD:

Comparing equation (vii) and equation (viii):

$\angle \text{BC'D} = \angle \text{BCD}$

Now, consider $\triangle \text{BC'C}$. The angle $\angle \text{BCD}$ is an exterior angle to $\triangle \text{BC'C}$ at vertex C (since C is between D and C').

By the Exterior Angle Theorem, the exterior angle of a triangle is equal to the sum of the two interior opposite angles.

$\angle \text{BCD} = \angle \text{BC'C} + \angle \text{C'BC}$

Since C' lies on the extension of DC, $\angle \text{BC'C}$ is the same angle as $\angle \text{BC'D}$.

So, $\angle \text{BCD} = \angle \text{BC'D} + \angle \text{C'BC}$

We found that $\angle \text{BC'D} = \angle \text{BCD}$. Substitute this into the equation:

$\angle \text{BCD} = \angle \text{BCD} + \angle \text{C'BC}$

Subtracting $\angle \text{BCD}$ from both sides:

$0^\circ = \angle \text{C'BC}$

This implies $\angle \text{C'BC} = 0^\circ$. Similar to Case 1, this means points C', B, and C are collinear. However, C' and B are distinct points on the circle through A, B, D, and C is a point on the line DC in the interior of the circle. This collinearity contradicts the fact that ABCD is a quadrilateral unless C coincides with B or C coincides with C'. C cannot coincide with B unless B is on the line DC. C cannot coincide with C' unless C is on the circle.

Therefore, our assumption that C lies inside the circle must also be false.

Since C cannot lie in the exterior and cannot lie in the interior of the circle passing through A, B, and D, the only remaining possibility is that C lies on the circle.

Thus, the points A, B, C, and D are concyclic, and the quadrilateral ABCD is cyclic.

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Exterior Angle of a Cyclic Quadrilateral

When one side of a cyclic quadrilateral is extended outwards, it forms an angle with the adjacent side of the quadrilateral. This angle, lying outside the boundary of the quadrilateral, is called an exterior angle of the cyclic quadrilateral.

In the illustration above, ABCD is a cyclic quadrilateral. If we extend the side AB to a point E, the angle formed, $\angle \text{CBE}$, is an exterior angle. For any vertex, there is a corresponding interior angle and an exterior angle formed by extending one of the sides meeting at that vertex.

The angle $\angle \text{ADC}$ (or $\angle \text{D}$) is the interior opposite angle to the exterior angle $\angle \text{CBE}$. The interior opposite angle is the interior angle at the vertex opposite to the vertex where the exterior angle is formed.

There is a specific property relating the exterior angle of a cyclic quadrilateral to its interior opposite angle.

Theorem: If a side of a cyclic quadrilateral is produced, then the exterior angle so formed is equal to the interior opposite angle.

This means that if ABCD is a cyclic quadrilateral and side AB is extended to E, then the exterior angle $\angle \text{CBE}$ is equal to the interior angle $\angle \text{ADC}$.

Given: A cyclic quadrilateral ABCD inscribed in a circle. The side AB is produced to a point E, forming the exterior angle $\angle \text{CBE}$.

To Prove: Exterior $\angle \text{CBE} = $ Interior opposite $\angle \text{ADC}$.

Proof:

Since ABCD is a cyclic quadrilateral, we know from the property of cyclic quadrilaterals that the sum of opposite interior angles is $180^\circ$ (they are supplementary).

Now consider the straight line ABE. The interior angle $\angle \text{ABC}$ and the exterior angle $\angle \text{CBE}$ are adjacent angles on a straight line. Angles that form a straight line sum up to $180^\circ$ (they form a linear pair).

We now have two equations where the sum of two angles is $180^\circ$:

Comparing equations (i) and (ii), since both sums are equal to $180^\circ$, we can equate them:

$\angle \text{ABC} + \angle \text{ADC} = \angle \text{ABC} + \angle \text{CBE}$

Subtracting $\angle \text{ABC}$ from both sides of the equation:

$\angle \text{ADC} = \angle \text{CBE}$

This shows that the exterior angle $\angle \text{CBE}$ is equal in measure to the interior opposite angle $\angle \text{ADC}$.

Hence, the theorem is proved.

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