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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
Chapter 8 Quadrilaterals (Concepts)
This chapter significantly advances our study of polygons by providing a more formal, rigorous, and proof-based exploration of Quadrilaterals, building upon the introductory concepts from Class 8 and leveraging our understanding of lines, angles, and crucially, triangle congruence. While we revisit the basic definition of a quadrilateral as a four-sided polygon, and recall its fundamental Angle Sum Property (the sum of the four interior angles is always $360^\circ$), our primary focus shifts towards a deeper understanding of specific types, especially parallelograms, and proving their characteristic properties using logical deduction.
The cornerstone of this chapter is the detailed investigation of Parallelograms – quadrilaterals defined by having both pairs of opposite sides parallel. We move beyond simply stating their properties to rigorously proving them, primarily using the SSS, SAS, ASA, or RHS congruence criteria for triangles. Key theorems regarding parallelograms that we will establish through proof include:
- Theorem 1: A diagonal divides a parallelogram into two congruent triangles.
- Theorem 2: In a parallelogram, opposite sides are equal in length.
- Theorem 3: In a parallelogram, opposite angles are equal in measure.
- Theorem 4: The diagonals of a parallelogram bisect each other (meaning they intersect at their midpoints).
Equally important are the converses of these properties, which provide the necessary and sufficient conditions to definitively prove that a given quadrilateral is, in fact, a parallelogram. We will establish theorems stating that a quadrilateral is a parallelogram if:
- Both pairs of opposite sides are equal, OR
- Both pairs of opposite angles are equal, OR
- The diagonals bisect each other, OR
- One pair of opposite sides is both equal AND parallel.
Mastering the application of these conditions in proofs is a central objective.
Having established the foundational properties of parallelograms, we then examine special types – rectangles, rhombuses, and squares – demonstrating how their unique properties are logical consequences of the general parallelogram properties combined with their specific defining characteristics. For instance:
- A Rectangle is defined as a parallelogram with at least one right angle. From this, we prove that all its angles must be right angles and, significantly, that its diagonals are equal in length.
- A Rhombus is defined as a parallelogram with all four sides equal (or equivalently, two equal adjacent sides). We prove from this that its diagonals bisect each other at right angles ($90^\circ$) and also bisect the angles of the rhombus.
- A Square is elegantly defined as a parallelogram that is both a rectangle and a rhombus. Consequently, it inherits all properties of both: four equal sides, four right angles, equal diagonals, and diagonals that bisect each other at right angles.
The chapter culminates with the introduction and proof of the extremely useful Mid-point Theorem concerning triangles, often proved using properties of parallelograms or triangle congruence. This theorem states: The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half the length of the third side. We also explore its important converse: The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side. These theorems provide powerful tools for solving a variety of geometric problems involving lengths, parallelism, and midpoints within triangles and related figures.
Quadrilateral and Terms Related to Quadrilaterals
In the previous chapter, we studied triangles, which are polygons with three sides. A quadrilateral is a closed plane figure (a polygon) formed by four line segments. It has four sides, four vertices, and four interior angles.
Let's consider a quadrilateral ABCD.
Terms Related to a Quadrilateral
Sides:
The four line segments that form the boundary of the quadrilateral are called its sides. In quadrilateral ABCD shown above, the sides are $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$.
Vertices:
The four points where adjacent sides meet are called its vertices. In quadrilateral ABCD, the vertices are A, B, C, and D.
Angles:
The interior angles formed at the vertices by the adjacent sides are called the angles of the quadrilateral. In quadrilateral ABCD, the angles are $\angle \text{A}$ (or $\angle \text{DAB}$), $\angle \text{B}$ (or $\angle \text{ABC}$), $\angle \text{C}$ (or $\angle \text{BCD}$), and $\angle \text{D}$ (or $\angle \text{CDA}$).
Diagonals:
The line segments connecting two non-adjacent vertices are called the diagonals. A quadrilateral has two diagonals. In quadrilateral ABCD, the diagonals are $\overline{AC}$ and $\overline{BD}$.
Adjacent Sides:
Two sides of a quadrilateral are adjacent if they share a common vertex. In quadrilateral ABCD, the pairs of adjacent sides are $(\overline{AB}, \overline{BC})$, $(\overline{BC}, \overline{CD})$, $(\overline{CD}, \overline{DA})$, and $(\overline{DA}, \overline{AB})$.
Opposite Sides:
Two sides of a quadrilateral are opposite if they do not share a common vertex. In quadrilateral ABCD, the pairs of opposite sides are $(\overline{AB}, \overline{CD})$ and $(\overline{BC}, \overline{DA})$.
Adjacent Angles:
Two angles of a quadrilateral are adjacent if they share a common arm (side). In quadrilateral ABCD, the pairs of adjacent angles are $(\angle \text{A}, \angle \text{B})$, $(\angle \text{B}, \angle \text{C})$, $(\angle \text{C}, \angle \text{D})$, and $(\angle \text{D}, \angle \text{A})$.
Opposite Angles:
Two angles of a quadrilateral are opposite if they do not share a common arm (side). In quadrilateral ABCD, the pairs of opposite angles are $(\angle \text{A}, \angle \text{C})$ and $(\angle \text{B}, \angle \text{D})$.
Interior and Exterior:
A quadrilateral, being a closed figure in a plane, divides the plane into three distinct parts: the interior (the region enclosed by the quadrilateral), the exterior (the region outside the quadrilateral), and the boundary (the set of points forming the quadrilateral itself).
Angle Sum Property of a Quadrilateral
Just as a triangle has an angle sum property (the sum of its interior angles is $180^\circ$), a quadrilateral also has a property regarding the sum of its interior angles.
Theorem 8.1: Angle Sum Property of a Quadrilateral
Theorem 8.1. The sum of the four interior angles of a quadrilateral is $360^\circ$.
Proof:
Given:
A quadrilateral ABCD.
To Prove:
$\angle \text{A} + \angle \text{B} + \angle \text{C} + \angle \text{D} = 360^\circ$.
Construction:
Draw a diagonal, say AC. This diagonal divides the quadrilateral ABCD into two triangles: $\triangle \text{ABC}$ and $\triangle \text{ADC}$.
Proof:
In $\triangle \text{ABC}$, by the Angle Sum Property of a triangle:
$\angle \text{BAC} + \angle \text{ABC} + \angle \text{BCA} = 180^\circ$
... (1)
In $\triangle \text{ADC}$, by the Angle Sum Property of a triangle:
$\angle \text{DAC} + \angle \text{ADC} + \angle \text{DCA} = 180^\circ$
... (2)
Adding equation (1) and equation (2), we get:
$(\angle \text{BAC} + \ $$ \angle \text{ABC} + \ $$ \angle \text{BCA}) + \ $$ (\angle \text{DAC} + \ $$ \angle \text{ADC} + \ $$ \angle \text{DCA}) = \ $$ 180^\circ + \ $$ 180^\circ$
$\angle \text{BAC} + \angle \text{ABC} + \angle \text{BCA} + \angle \text{DAC} + \angle \text{ADC} + \angle \text{DCA} = 360^\circ$
Rearranging the terms and grouping the angles of the quadrilateral:
$(\angle \text{BAC} + \angle \text{DAC}) + \angle \text{ABC} + (\angle \text{BCA} + \angle \text{DCA}) + \angle \text{ADC} = 360^\circ$
From the figure, we can observe the following relationships:
$\angle \text{BAC} + \angle \text{DAC} = \angle \text{BAD} = \angle \text{A}$
$\angle \text{BCA} + \angle \text{DCA} = \angle \text{BCD} = \angle \text{C}$
Substituting these sums back into the equation, we get:
$\angle \text{A} + \angle \text{B} + \angle \text{C} + \angle \text{D} = 360^\circ$
Thus, the sum of the interior angles of a quadrilateral is $360^\circ$. This property is known as the Angle Sum Property of a Quadrilateral.
Example 1. The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.
Answer:
Let the four angles of the quadrilateral be $3x^\circ$, $5x^\circ$, $9x^\circ$, and $13x^\circ$, where $x$ is a positive number.
By the Angle Sum Property of a quadrilateral, the sum of these angles must be $360^\circ$.
$3x + 5x + 9x + 13x = 360$
Combining the terms on the left side:
$(3+5+9+13)x = 360$
$30x = 360$
To find the value of $x$, divide both sides of the equation by 30:
$x = \frac{360}{30}$
$x = 12$
Now, we can find the measure of each angle using the value of $x$:
- First angle = $3x^\circ = 3 \times 12^\circ = 36^\circ$
- Second angle = $5x^\circ = 5 \times 12^\circ = 60^\circ$
- Third angle = $9x^\circ = 9 \times 12^\circ = 108^\circ$
- Fourth angle = $13x^\circ = 13 \times 12^\circ = 156^\circ$
Thus, the four angles of the quadrilateral are $36^\circ$, $60^\circ$, $108^\circ$, and $156^\circ$.
To verify, we can add the angles:
$36^\circ + 60^\circ + 108^\circ + 156^\circ = 96^\circ + 108^\circ + 156^\circ$
$ = 204^\circ + 156^\circ = 360^\circ$
The sum is $360^\circ$, which confirms our answer is correct based on the Angle Sum Property.
Types of Quadrilaterals
Quadrilaterals are four-sided polygons. They are classified into different types based on the specific properties of their sides, angles, and diagonals. Understanding these classifications is crucial for studying their unique characteristics.
1. Trapezium
A trapezium is a quadrilateral where at least one pair of opposite sides is parallel.
In the quadrilateral ABCD shown below, if side AB is parallel to side CD ($\text{AB} \parallel \text{CD}$), then ABCD is a trapezium. The parallel sides (AB and CD) are called the bases, and the non-parallel sides (AD and BC) are called the legs.
Isosceles Trapezium
A trapezium is called an isosceles trapezium if its non-parallel sides (legs) are equal in length.
Properties of an isosceles trapezium include:
- The legs are equal ($\text{AD} = \text{BC}$).
- The base angles are equal ($\angle \text{DAB} = \angle \text{CBA}$ and $\angle \text{ADC} = \angle \text{BCD}$).
- The diagonals are equal in length ($\text{AC} = \text{BD}$).
2. Kite
A kite is a quadrilateral that has two distinct pairs of equal-length adjacent sides.
In the figure, quadrilateral ABCD is a kite because it has two pairs of equal adjacent sides: $\text{AB} = \text{AD}$ and $\text{CB} = \text{CD}$.
Key properties of a kite:
- The diagonals are perpendicular to each other ($\text{AC} \perp \text{BD}$).
- One of the diagonals (the main diagonal, AC) is the perpendicular bisector of the other diagonal (BD).
- One pair of opposite angles is equal (the angles between the unequal sides, i.e., $\angle \text{ABC} = \angle \text{ADC}$).
- The main diagonal (AC) bisects the angles at the vertices it joins ($\angle \text{DAB}$ and $\angle \text{BCD}$).
3. Parallelogram
A parallelogram is a quadrilateral where both pairs of opposite sides are parallel.
In quadrilateral ABCD, if $\text{AB} \parallel \text{CD}$ and $\text{AD} \parallel \text{BC}$, then ABCD is a parallelogram. This is a very important type of quadrilateral, and several other shapes are special cases of a parallelogram.
The fundamental properties of a parallelogram will be proven later, but they include:
- Opposite sides are equal in length.
- Opposite angles are equal.
- The diagonals bisect each other.
Special Types of Parallelograms
Rhombus, Rectangle, and Square are all special types of parallelograms, meaning they have all the properties of a parallelogram plus some extra ones.
Rhombus
A rhombus is a parallelogram with all four sides of equal length.
So, a rhombus has $\text{AB} = \text{BC} = \text{CD} = \text{DA}$.
In addition to the properties of a parallelogram, a rhombus has:
- All four sides are equal.
- The diagonals bisect each other at right angles ($90^\circ$).
- The diagonals bisect the angles of the rhombus.
Rectangle
A rectangle is a parallelogram with one interior angle being a right angle ($90^\circ$).
If one angle is $90^\circ$, then all four angles must be $90^\circ$ because consecutive angles in a parallelogram are supplementary ($180^\circ - 90^\circ = 90^\circ$) and opposite angles are equal.
In addition to the properties of a parallelogram, a rectangle has:
- All four interior angles are right angles.
- The diagonals are equal in length.
Square
A square is the most specific type of parallelogram. It can be defined as:
- A rhombus that is also a rectangle.
- A rectangle that is also a rhombus.
This means a square is a parallelogram with four equal sides and four right angles. It possesses all the properties of a parallelogram, a rhombus, and a rectangle combined.
Key properties of a square:
- All four sides are equal.
- All four angles are $90^\circ$.
- Diagonals are equal, bisect each other at right angles, and bisect the vertex angles (into $45^\circ$ angles).
Hierarchy of Quadrilaterals
The different types of quadrilaterals are related in a hierarchical structure. Understanding this helps to see how properties are inherited from more general shapes to more specific ones.
Key relationships to remember:
- A Square is a special type of Rectangle and also a special type of Rhombus.
- Both Rectangles and Rhombuses are special types of Parallelograms.
- A Parallelogram is a special type of Trapezium (since it has at least one pair of parallel sides).
- A Kite is a distinct category, but a rhombus is also a kite (one where all adjacent sides are equal).
The following table summarizes the key properties of these quadrilaterals:
| Property | Parallelogram | Rectangle | Rhombus | Square | Kite |
|---|---|---|---|---|---|
| Opposite sides are parallel | Yes (both pairs) | Yes (both pairs) | Yes (both pairs) | Yes (both pairs) | No |
| Opposite sides are equal | Yes | Yes | Yes | Yes | No |
| All sides are equal | No | No | Yes | Yes | No |
| Opposite angles are equal | Yes | Yes | Yes | Yes | Only one pair |
| All angles are right angles | No | Yes | No | Yes | No |
| Diagonals bisect each other | Yes | Yes | Yes | Yes | Only one is bisected |
| Diagonals are equal | No | Yes | No | Yes | No |
| Diagonals are perpendicular | No | No | Yes | Yes | Yes |
Properties of a Parallelogram
A parallelogram is a special type of quadrilateral with specific properties related to its sides, angles, and diagonals. These properties are fundamental to understanding parallelograms and can be proven using the concepts of triangle congruence and properties of parallel lines.
Theorem 8.1: Diagonal Divides into Congruent Triangles
Theorem 8.1. A diagonal of a parallelogram divides it into two congruent triangles.
Proof:
Given:
A parallelogram ABCD. Let AC be a diagonal.
To Prove:
$\triangle \text{ABC} \cong \triangle \text{CDA}$.
Construction:
Draw the diagonal AC.
Proof:
Consider $\triangle \text{ABC}$ and $\triangle \text{CDA}$.
Since ABCD is a parallelogram, we know that AB $\parallel$ DC and AD $\parallel$ BC by definition.
Considering transversal AC intersecting the parallel lines AB and DC:
$\angle \text{BAC} = \angle \text{DCA}$
(Alternate Interior Angles) ... (1)
Considering transversal AC intersecting the parallel lines AD and BC:
$\angle \text{BCA} = \angle \text{DAC}$
(Alternate Interior Angles) ... (2)
Also, the side AC is common to both triangles $\triangle \text{ABC}$ and $\triangle \text{CDA}$.
$\text{AC} = \text{CA}$
(Common side) ... (3)
Now, let's compare the two triangles $\triangle \text{ABC}$ and $\triangle \text{CDA}$:
- $\angle \text{BAC} = \angle \text{DCA}$ (From equation (1))
- $\text{AC} = \text{CA}$ (From equation (3))
- $\angle \text{BCA} = \angle \text{DAC}$ (From equation (2))
By the Angle-Side-Angle (ASA) congruence rule, if two angles and the included side of one triangle are equal to two angles and the included side of another triangle, then the triangles are congruent.
$\triangle \text{ABC} \cong \triangle \text{CDA}$
Thus, a diagonal of a parallelogram divides it into two congruent triangles. This completes the proof.
Properties Derived from Theorem 8.1
Since the diagonal divides the parallelogram into two congruent triangles, the corresponding parts of these congruent triangles (CPCTC - Corresponding Parts of Congruent Triangles) must be equal. This leads to important properties of parallelograms.
Theorem 8.2: Opposite Sides are Equal
Theorem 8.2. In a parallelogram, opposite sides are equal in length.
Proof:
Given:
Parallelogram ABCD.
To Prove:
$\text{AB} = \text{CD}$ and $\text{BC} = \text{DA}$.
Proof:
Draw diagonal AC. From Theorem 8.1, we know that a diagonal of a parallelogram divides it into two congruent triangles. Therefore,
$\triangle \text{ABC} \cong \triangle \text{CDA}$
By the property of Corresponding Parts of Congruent Triangles (CPCTC), the corresponding sides of these congruent triangles are equal.
$\text{AB} = \text{CD}$
(CPCTC)
$\text{BC} = \text{DA}$
(CPCTC)
Thus, in a parallelogram, both pairs of opposite sides are equal.
Theorem 8.3: Opposite Angles are Equal
Theorem 8.3. In a parallelogram, opposite angles are equal in measure.
Proof:
Given:
Parallelogram ABCD.
To Prove:
$\angle \text{A} = \angle \text{C}$ and $\angle \text{B} = \angle \text{D}$.
Proof:
Draw diagonal AC. From Theorem 8.1, we know that
$\triangle \text{ABC} \cong \triangle \text{CDA}$
By CPCTC, the corresponding angles of these congruent triangles are equal. This directly gives us one pair of opposite angles:
$\angle \text{ABC} = \angle \text{CDA}$
(CPCTC, i.e., $\angle \text{B} = \angle \text{D}$)
For the other pair of opposite angles, $\angle \text{A}$ and $\angle \text{C}$, we look at the angles formed by the diagonal:
$\angle \text{A} = \angle \text{BAD} = \angle \text{BAC} + \angle \text{DAC}$
$\angle \text{C} = \angle \text{BCD} = \angle \text{BCA} + \angle \text{DCA}$
From the congruence $\triangle \text{ABC} \cong \triangle \text{CDA}$ (Theorem 8.1 proof), we had:
$\angle \text{BAC} = \angle \text{DCA}$
(Alternate Interior Angles)
$\angle \text{DAC} = \angle \text{BCA}$
(Alternate Interior Angles)
Adding these two equalities:
$\angle \text{BAC} + \angle \text{DAC} = \angle \text{DCA} + \angle \text{BCA}$
By angle addition, the left side is $\angle \text{BAD}$ ($\angle \text{A}$) and the right side is $\angle \text{BCD}$ ($\angle \text{C}$).
$\angle \text{A} = \angle \text{C}$
Thus, in a parallelogram, both pairs of opposite angles are equal ($\angle \text{A} = \angle \text{C}$ and $\angle \text{B} = \angle \text{D}$).
Consecutive Angles are Supplementary
In a parallelogram, consecutive angles (angles that share a common side) are supplementary, meaning their sum is $180^\circ$. This property arises directly from the definition of parallel lines.
Consider parallelogram ABCD with AB $\parallel$ DC. AD is a transversal. By the property of interior angles on the same side of a transversal, the sum of consecutive interior angles is $180^\circ$.
$\angle \text{A} + \angle \text{D} = 180^\circ$
Similarly, considering AD $\parallel$ BC and transversal AB, AD $\parallel$ BC and transversal CD, and AB $\parallel$ DC and transversal BC, we get:
$\angle \text{A} + \angle \text{B} = 180^\circ$
$\angle \text{B} + \angle \text{C} = 180^\circ$
$\angle \text{C} + \angle \text{D} = 180^\circ$
This can also be confirmed using the Angle Sum Property ($\angle \text{A} + \angle \text{B} + \angle \text{C} + \angle \text{D} = 360^\circ$) and Theorem 8.3 ($\angle \text{A} = \angle \text{C}$, $\angle \text{B} = \angle \text{D}$). Substituting $\angle \text{C} = \angle \text{A}$ and $\angle \text{D} = \angle \text{B}$ into the sum of angles equation:
$\angle \text{A} + \angle \text{B} + \angle \text{A} + \angle \text{B} = 360^\circ$
$2\angle \text{A} + 2\angle \text{B} = 360^\circ$
Dividing by 2:
$\angle \text{A} + \angle \text{B} = 180^\circ$
Similarly, other pairs of consecutive angles are supplementary.
Theorem 8.4: Diagonals Bisect Each Other
Theorem 8.4. The diagonals of a parallelogram bisect each other.
Proof:
Given:
Parallelogram ABCD. Diagonals AC and BD intersect at point O.
To Prove:
The diagonals bisect each other, i.e., $\text{AO} = \text{CO}$ and $\text{BO} = \text{DO}$.
Construction:
Draw diagonals AC and BD, intersecting at point O.
Proof:
Consider $\triangle \text{AOB}$ and $\triangle \text{COD}$.
Since ABCD is a parallelogram, AB $\parallel$ DC.
Considering transversal AC intersecting parallel lines AB and DC:
$\angle \text{OAB} = \angle \text{OCD}$
(Alternate Interior Angles) ... (1)
(Note that $\angle \text{OAB}$ is the same as $\angle \text{CAB}$ or $\angle \text{BAC}$, and $\angle \text{OCD}$ is the same as $\angle \text{ACD}$ or $\angle \text{DCA}$)
Considering transversal BD intersecting parallel lines AB and DC:
$\angle \text{OBA} = \angle \text{ODC}$
(Alternate Interior Angles) ... (2)
(Note that $\angle \text{OBA}$ is the same as $\angle \text{DBA}$ or $\angle \text{ABD}$, and $\angle \text{ODC}$ is the same as $\angle \text{BDC}$ or $\angle \text{CD B}$)
Also, we know from Theorem 8.2 that opposite sides of a parallelogram are equal:
$\text{AB} = \text{CD}$
(Opposite sides of parallelogram) ... (3)
Now, consider $\triangle \text{AOB}$ and $\triangle \text{COD}$ again. We have:
- $\angle \text{OAB} = \angle \text{OCD}$ (From equation (1))
- $\text{AB} = \text{CD}$ (From equation (3))
- $\angle \text{OBA} = \angle \text{ODC}$ (From equation (2))
By the Angle-Side-Angle (ASA) congruence rule:
$\triangle \text{AOB} \cong \triangle \text{COD}$
By CPCTC, their corresponding parts are equal:
$\text{AO} = \text{CO}$
(CPCTC)
$\text{BO} = \text{DO}$
(CPCTC)
This shows that the diagonals AC and BD bisect each other at point O. This completes the proof.
Conditions for a Quadrilateral to be a Parallelogram (Converse Theorems)
The following theorems are converses of the properties we just proved. They provide conditions under which a quadrilateral can be guaranteed to be a parallelogram.
Theorem 8.5: If opposite sides are equal...
Theorem 8.5. If in a quadrilateral, each pair of opposite sides is equal, then it is a parallelogram.
Proof:
Given:
A quadrilateral ABCD such that $\text{AB} = \text{CD}$ and $\text{BC} = \text{DA}$.
To Prove:
ABCD is a parallelogram (i.e., AB $\parallel$ DC and BC $\parallel$ DA).
Construction:
Draw diagonal AC.
Proof:
Consider $\triangle \text{ABC}$ and $\triangle \text{CDA}$.
We are given that the opposite sides are equal:
$\text{AB} = \text{CD}$
(Given)
$\text{BC} = \text{DA}$
(Given)
The diagonal AC is common to both triangles:
$\text{AC} = \text{CA}$
(Common side)
By the Side-Side-Side (SSS) congruence rule, if the three sides of one triangle are equal to the three corresponding sides of another triangle, then the triangles are congruent.
$\triangle \text{ABC} \cong \triangle \text{CDA}$
By CPCTC, the corresponding angles are equal:
$\angle \text{BAC} = \angle \text{DCA}$
(CPCTC) ... (1)
$\angle \text{BCA} = \angle \text{DAC}$
(CPCTC) ... (2)
In equation (1), $\angle \text{BAC}$ and $\angle \text{DCA}$ are alternate interior angles formed by transversal AC intersecting lines AB and DC. Since these alternate interior angles are equal, it implies that AB $\parallel$ DC.
In equation (2), $\angle \text{BCA}$ and $\angle \text{DAC}$ are alternate interior angles formed by transversal AC intersecting lines BC and DA. Since these alternate interior angles are equal, it implies that BC $\parallel$ DA.
Since both pairs of opposite sides are parallel (AB $\parallel$ DC and BC $\parallel$ DA), by definition, ABCD is a parallelogram.
Theorem 8.6: If opposite angles are equal...
Theorem 8.6. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Proof:
Given:
A quadrilateral ABCD such that $\angle \text{A} = \angle \text{C}$ and $\angle \text{B} = \angle \text{D}$.
To Prove:
ABCD is a parallelogram (i.e., AB $\parallel$ DC and BC $\parallel$ DA).
Proof:
The sum of the interior angles of any quadrilateral is $360^\circ$ (Angle Sum Property of a Quadrilateral).
$\angle \text{A} + \angle \text{B} + \angle \text{C} + \angle \text{D} = 360^\circ$
We are given $\angle \text{A} = \angle \text{C}$ and $\angle \text{B} = \angle \text{D}$. Substitute $\angle \text{C}$ with $\angle \text{A}$ and $\angle \text{D}$ with $\angle \text{B}$ in the equation:
$\angle \text{A} + \angle \text{B} + \angle \text{A} + \angle \text{B} = 360^\circ$
$2\angle \text{A} + 2\angle \text{B} = 360^\circ$
Dividing the entire equation by 2:
$\angle \text{A} + \angle \text{B} = 180^\circ$
... (1)
Similarly, by substituting $\angle \text{A} = \angle \text{C}$ and $\angle \text{B} = \angle \text{D}$ in different combinations into the sum of angles equation, we can show:
$\angle \text{B} + \angle \text{C} = 180^\circ$
$\angle \text{C} + \angle \text{D} = 180^\circ$
$\angle \text{D} + \angle \text{A} = 180^\circ$
Consider lines AD and BC and transversal AB. The angles $\angle \text{A}$ and $\angle \text{B}$ are consecutive interior angles. From equation (1), their sum is $180^\circ$. Since the sum of consecutive interior angles is $180^\circ$, by the Converse of the Interior Angles on the Same Side of a Transversal Theorem, the lines AD and BC must be parallel.
AD $\parallel$ BC
Similarly, consider lines AB and DC and transversal AD. The angles $\angle \text{A}$ and $\angle \text{D}$ are consecutive interior angles. Since $\angle \text{A} + \angle \text{D} = 180^\circ$, by the Converse of the Interior Angles Theorem, the lines AB and DC must be parallel.
AB $\parallel$ DC
Since both pairs of opposite sides are parallel (AD $\parallel$ BC and AB $\parallel$ DC), by definition, ABCD is a parallelogram.
Theorem 8.7: If diagonals bisect each other...
Theorem 8.7. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Proof:
Given:
A quadrilateral ABCD. Diagonals AC and BD intersect at point O such that $\text{AO} = \text{CO}$ and $\text{BO} = \text{DO}$.
To Prove:
ABCD is a parallelogram (i.e., AB $\parallel$ DC and BC $\parallel$ DA).
Proof:
Consider $\triangle \text{AOB}$ and $\triangle \text{COD}$.
We are given that the diagonals bisect each other:
$\text{AO} = \text{CO}$
(Given) ... (1)
$\text{BO} = \text{DO}$
(Given) ... (2)
The angles $\angle \text{AOB}$ and $\angle \text{COD}$ are vertically opposite angles:
$\angle \text{AOB} = \angle \text{COD}$
(Vertically Opposite Angles) ... (3)
By the Side-Angle-Side (SAS) congruence rule, if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the triangles are congruent.
$\triangle \text{AOB} \cong \triangle \text{COD}$
(By SAS, using (1), (3), (2))
By CPCTC, the corresponding parts are equal:
$\text{AB} = \text{CD}$
(CPCTC) ... (4)
$\angle \text{OAB} = \angle \text{OCD}$
(CPCTC) ... (5)
In equation (5), $\angle \text{OAB}$ (which is $\angle \text{BAC}$) and $\angle \text{OCD}$ (which is $\angle \text{DCA}$) are alternate interior angles formed by transversal AC intersecting lines AB and DC. Since these alternate interior angles are equal, it implies that AB $\parallel$ DC.
AB $\parallel$ DC
From equation (4), we have $\text{AB} = \text{CD}$. So, we have shown that one pair of opposite sides (AB and CD) is both equal and parallel. This is a sufficient condition for a quadrilateral to be a parallelogram, as stated in Theorem 8.8.
Alternatively, we could also prove that AD $\parallel$ BC and AD = BC by considering $\triangle \text{AOD}$ and $\triangle \text{COB}$ (also congruent by SAS using $\text{AO}=\text{CO}$, $\text{DO}=\text{BO}$, and $\angle \text{AOD} = \angle \text{COB}$). Since both pairs of opposite sides are parallel, ABCD is a parallelogram.
Theorem 8.8: If one pair of opposite sides is equal and parallel...
Theorem 8.8. A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
Proof:
Given:
A quadrilateral ABCD such that $\text{AB} \parallel \text{DC}$ and $\text{AB} = \text{DC}$.
To Prove:
ABCD is a parallelogram (i.e., AD $\parallel$ BC).
Construction:
Draw diagonal AC.
Proof:
Consider $\triangle \text{ABC}$ and $\triangle \text{CDA}$.
We are given that AB $\parallel$ DC and AB = DC.
$\text{AB} = \text{DC}$
(Given) ... (1)
Since AB $\parallel$ DC and AC is a transversal, the alternate interior angles are equal:
$\angle \text{BAC} = \angle \text{DCA}$
(Alternate Interior Angles) ... (2)
The side AC is common to both triangles:
$\text{AC} = \text{CA}$
(Common side) ... (3)
Now, let's compare $\triangle \text{ABC}$ and $\triangle \text{CDA}$:
- $\text{AB} = \text{DC}$ (From equation (1))
- $\angle \text{BAC} = \angle \text{DCA}$ (From equation (2))
- $\text{AC} = \text{CA}$ (From equation (3))
By the Side-Angle-Side (SAS) congruence rule:
$\triangle \text{ABC} \cong \triangle \text{CDA}$
(By SAS, using (1), (2), (3))
By CPCTC, the corresponding angles are equal:
$\angle \text{BCA} = \angle \text{DAC}$
(CPCTC)
These angles, $\angle \text{BCA}$ and $\angle \text{DAC}$, are alternate interior angles formed by transversal AC intersecting lines BC and DA. Since these alternate interior angles are equal, it implies that BC $\parallel$ DA.
BC $\parallel$ DA
Since both pairs of opposite sides are parallel (AB $\parallel$ DC is given, and we proved BC $\parallel$ DA), by definition, ABCD is a parallelogram.
Properties of a Rectangle
A rectangle is defined as a parallelogram with one interior angle measuring $90^\circ$. Because a rectangle is a type of parallelogram, it possesses all the properties of a parallelogram, in addition to some specific properties of its own.
Properties Inherited from Parallelograms
Since every rectangle is a parallelogram, the following properties hold true for any rectangle ABCD:
- Opposite sides are parallel: $\text{AB} \parallel \text{DC}$ and $\text{BC} \parallel \text{DA}$.
- Opposite sides are equal: $\text{AB} = \text{DC}$ and $\text{BC} = \text{DA}$.
- Opposite angles are equal: $\angle \text{A} = \angle \text{C}$ and $\angle \text{B} = \angle \text{D}$.
- Consecutive angles are supplementary: $\angle \text{A} + \angle \text{B} = 180^\circ$, $\angle \text{B} + \angle \text{C} = 180^\circ$, $\angle \text{C} + \angle \text{D} = 180^\circ$, $\angle \text{D} + \angle \text{A} = 180^\circ$.
- Diagonals bisect each other: The point where the diagonals intersect is the midpoint of both diagonals.
Additional Properties Specific to a Rectangle
All angles are $90^\circ$ (Right Angles):
The definition of a rectangle states that at least one angle is $90^\circ$. Let's prove that this implies all four angles are $90^\circ$.
Assume that in parallelogram ABCD, $\angle \text{A} = 90^\circ$.
Since ABCD is a parallelogram, consecutive angles are supplementary. So, $\angle \text{A} + \angle \text{B} = 180^\circ$.
$90^\circ + \angle \text{B} = 180^\circ$
Subtracting $90^\circ$ from both sides:
$\angle \text{B} = 180^\circ - 90^\circ$
$\angle \text{B} = 90^\circ$
Also, in a parallelogram, opposite angles are equal. So,
$\angle \text{C} = \angle \text{A}$
(Opposite angles of parallelogram)
$\angle \text{C} = 90^\circ$
And,
$\angle \text{D} = \angle \text{B}$
(Opposite angles of parallelogram)
$\angle \text{D} = 90^\circ$
Thus, if one angle of a parallelogram is $90^\circ$, all four angles are $90^\circ$. Hence, every angle in a rectangle is a right angle.
Theorem 8.10: Diagonals of a rectangle are equal
Theorem 8.10. The diagonals of a rectangle are equal in length.
Proof:
Given:
Rectangle ABCD. AC and BD are its diagonals.
To Prove:
$\text{AC} = \text{BD}$.
Proof:
Consider the triangles $\triangle \text{ABC}$ and $\triangle \text{DCB}$.
Since ABCD is a rectangle, it is a parallelogram, so its opposite sides are equal:
$\text{AB} = \text{DC}$
(Opposite sides of parallelogram) ... (1)
The side BC is common to both triangles:
$\text{BC} = \text{CB}$
(Common side) ... (2)
Since ABCD is a rectangle, all its angles are $90^\circ$. Therefore,
$\angle \text{ABC} = 90^\circ$
$\angle \text{DCB} = 90^\circ$
So,
$\angle \text{ABC} = \angle \text{DCB}$
($90^\circ$ each) ... (3)
Now, compare $\triangle \text{ABC}$ and $\triangle \text{DCB}$:
- $\text{AB} = \text{DC}$ (From equation (1))
- $\angle \text{ABC} = \angle \text{DCB}$ (From equation (3))
- $\text{BC} = \text{CB}$ (From equation (2))
By the Side-Angle-Side (SAS) congruence rule:
$\triangle \text{ABC} \cong \triangle \text{DCB}$
(By SAS, using (1), (3), (2))
By CPCTC, their corresponding parts are equal. The diagonals AC and BD are the hypotenuses of these right-angled triangles.
$\text{AC} = \text{DB}$
(CPCTC)
Thus, the diagonals of a rectangle are equal.
Conditions for a Parallelogram to be a Rectangle (Converse Theorems)
The following theorems provide conditions under which a parallelogram can be classified as a rectangle.
Theorem 8.9: If one angle is $90^\circ$...
Theorem 8.9. If in a parallelogram one angle is a right angle, then it is a rectangle.
Proof:
Given:
A parallelogram ABCD with one angle equal to $90^\circ$. Let $\angle \text{A} = 90^\circ$.
To Prove:
ABCD is a rectangle (i.e., all angles are $90^\circ$).
Proof:
Since ABCD is a parallelogram, opposite angles are equal ($\angle \text{C} = \angle \text{A}$) and consecutive angles are supplementary ($\angle \text{A} + \angle \text{B} = 180^\circ$, $\angle \text{B} + \angle \text{C} = 180^\circ$, etc.).
Given $\angle \text{A} = 90^\circ$.
Since $\angle \text{C} = \angle \text{A}$, we have $\angle \text{C} = 90^\circ$.
Since $\angle \text{A} + \angle \text{B} = 180^\circ$, we have $90^\circ + \angle \text{B} = 180^\circ$, which implies $\angle \text{B} = 90^\circ$.
Since $\angle \text{D} = \angle \text{B}$, we have $\angle \text{D} = 90^\circ$.
Since all four angles ($\angle \text{A}, \angle \text{B}, \angle \text{C}, \angle \text{D}$) are equal to $90^\circ$, and ABCD is a parallelogram, it satisfies the definition of a rectangle. Thus, ABCD is a rectangle.
Theorem 8.11: If diagonals are equal...
Theorem 8.11. If the diagonals of a parallelogram are equal, then it is a rectangle.
Proof:
Given:
A parallelogram ABCD with diagonals AC and BD such that $\text{AC} = \text{BD}$.
To Prove:
ABCD is a rectangle (i.e., one of its angles is $90^\circ$).
Proof:
Consider the triangles $\triangle \text{ABC}$ and $\triangle \text{DCB}$.
Since ABCD is a parallelogram, opposite sides are equal:
$\text{AB} = \text{DC}$
(Opposite sides of parallelogram) ... (1)
The side BC is common to both triangles:
$\text{BC} = \text{CB}$
(Common side) ... (2)
We are given that the diagonals are equal:
$\text{AC} = \text{BD}$
(Given) ... (3)
Now, compare $\triangle \text{ABC}$ and $\triangle \text{DCB}$:
- $\text{AB} = \text{DC}$ (From equation (1))
- $\text{BC} = \text{CB}$ (From equation (2))
- $\text{AC} = \text{BD}$ (From equation (3))
By the Side-Side-Side (SSS) congruence rule:
$\triangle \text{ABC} \cong \triangle \text{DCB}$
(By SSS, using (1), (2), (3))
By CPCTC, the corresponding angles are equal:
$\angle \text{ABC} = \angle \text{DCB}$
(CPCTC) ... (4)
In parallelogram ABCD, $\angle \text{ABC}$ and $\angle \text{DCB}$ are consecutive interior angles on the same side of transversal BC. Therefore, they are supplementary.
$\angle \text{ABC} + \angle \text{DCB} = 180^\circ$
(Consecutive angles are supplementary) ... (5)
Substitute $\angle \text{DCB}$ with $\angle \text{ABC}$ from equation (4) into equation (5):
$\angle \text{ABC} + \angle \text{ABC} = 180^\circ$
$2\angle \text{ABC} = 180^\circ$
$\angle \text{ABC} = \frac{180^\circ}{2}$
$\angle \text{ABC} = 90^\circ$
Since ABCD is a parallelogram with one angle ($\angle \text{ABC}$) measuring $90^\circ$, by Theorem 8.9, it is a rectangle.
Properties of a Rhombus
A rhombus is defined as a parallelogram in which all four sides are equal in length. Since a rhombus is a special type of parallelogram, it inherits all the properties of a parallelogram, and it also has additional properties specific to its equal sides.
Properties Inherited from Parallelograms
As a rhombus is a parallelogram, the following properties are automatically true for any rhombus ABCD:
- Opposite sides are parallel: $\text{AB} \parallel \text{DC}$ and $\text{BC} \parallel \text{DA}$.
- Opposite sides are equal: $\text{AB} = \text{DC}$ and $\text{BC} = \text{DA}$. (This is consistent with the definition of a rhombus having all sides equal).
- Opposite angles are equal: $\angle \text{A} = \angle \text{C}$ and $\angle \text{B} = \angle \text{D}$.
- Consecutive angles are supplementary: $\angle \text{A} + \angle \text{B} = 180^\circ$, $\angle \text{B} + \angle \text{C} = 180^\circ$, $\angle \text{C} + \angle \text{D} = 180^\circ$, $\angle \text{D} + \angle \text{A} = 180^\circ$.
- Diagonals bisect each other: The point where the diagonals intersect is the midpoint of both diagonals.
Additional Properties Specific to a Rhombus
All sides are equal:
This property is part of the definition of a rhombus: $\text{AB} = \text{BC} = \text{CD} = \text{DA}$.
Theorem 8.12: Diagonals are Perpendicular Bisectors and Bisect Angles
Theorem 8.12. The diagonals of a rhombus are perpendicular bisectors of each other and bisect the angles of the rhombus.
Proof:
Given:
Rhombus ABCD. Diagonals AC and BD intersect at O.
To Prove:
1. Diagonals are perpendicular: $\text{AC} \perp \text{BD}$.
2. Diagonals bisect the angles: BD bisects $\angle \text{B}$ and $\angle \text{D}$, and AC bisects $\angle \text{A}$ and $\angle \text{C}$.
(Diagonals bisecting each other is already true since a rhombus is a parallelogram - Theorem 8.4).
Proof of Perpendicularity:
Since a rhombus is a parallelogram, its diagonals bisect each other. Thus, O is the midpoint of AC and BD.
$\text{AO} = \text{CO}$
(Diagonals of a parallelogram bisect each other) ... (1)
$\text{BO} = \text{DO}$
(Diagonals of a parallelogram bisect each other)
Consider $\triangle \text{AOD}$ and $\triangle \text{COD}$.
All sides of a rhombus are equal:
$\text{AD} = \text{CD}$
(Sides of a rhombus) ... (2)
The side OD is common to both triangles:
$\text{OD} = \text{OD}$
(Common side) ... (3)
Now, compare $\triangle \text{AOD}$ and $\triangle \text{COD}$:
- $\text{AO} = \text{CO}$ (From equation (1))
- $\text{AD} = \text{CD}$ (From equation (2))
- $\text{OD} = \text{OD}$ (From equation (3))
By the Side-Side-Side (SSS) congruence rule:
$\triangle \text{AOD} \cong \triangle \text{COD}$
(By SSS, using (1), (2), (3))
By CPCTC, corresponding angles are equal:
$\angle \text{AOD} = \angle \text{COD}$
(CPCTC) ... (4)
The angles $\angle \text{AOD}$ and $\angle \text{COD}$ form a linear pair on the straight line BD.
$\angle \text{AOD} + \angle \text{COD} = 180^\circ$
(Linear Pair Axiom)
Substitute $\angle \text{COD}$ with $\angle \text{AOD}$ from equation (4):
$\angle \text{AOD} + \angle \text{AOD} = 180^\circ$
$2\angle \text{AOD} = 180^\circ$
$\angle \text{AOD} = \frac{180^\circ}{2}$
$\angle \text{AOD} = 90^\circ$
Since $\angle \text{AOD} = 90^\circ$, the diagonals AC and BD intersect at right angles, i.e., AC $\perp$ BD.
Combined with the property that diagonals of a parallelogram bisect each other, this proves that the diagonals of a rhombus are perpendicular bisectors of each other.
Proof of Angle Bisection:
From the congruence $\triangle \text{AOD} \cong \triangle \text{COD}$, by CPCTC, we have:
$\angle \text{ADO} = \angle \text{CDO}$
(CPCTC)
This shows that the diagonal BD bisects $\angle \text{ADC}$ (which is $\angle \text{D}$).
Now consider $\triangle \text{AOB}$ and $\triangle \text{COB}$.
- $\text{AB} = \text{CB}$ (Sides of a rhombus)
- $\text{AO} = \text{CO}$ (Diagonals of a parallelogram bisect each other)
- $\text{BO} = \text{BO}$ (Common side)
By SSS congruence rule, $\triangle \text{AOB} \cong \triangle \text{COB}$.
By CPCTC:
$\angle \text{ABO} = \angle \text{CBO}$
(CPCTC)
This shows that the diagonal BD bisects $\angle \text{ABC}$ (which is $\angle \text{B}$).
Similarly, by considering $\triangle \text{AOB} \cong \triangle \text{AOD}$ (SAS congruence using AO=AO, $\angle \text{AOB}=\angle \text{AOD}=90^\circ$, BO=DO) we get $\angle \text{BAO} = \angle \text{DAO}$. This shows AC bisects $\angle \text{A}$.
And by considering $\triangle \text{COB} \cong \triangle \text{COD}$ (SAS congruence using CO=CO, $\angle \text{COB}=\angle \text{COD}=90^\circ$, BO=DO) we get $\angle \text{BCO} = \angle \text{DCO}$. This shows AC bisects $\angle \text{C}$.
Thus, the diagonals of a rhombus bisect the angles at the vertices.
Conditions for a Parallelogram to be a Rhombus (Converse Theorems)
The following theorems provide conditions under which a parallelogram can be classified as a rhombus.
Theorem 8.13: If diagonals are perpendicular...
Theorem 8.13. If the diagonals of a parallelogram are perpendicular to each other, then it is a rhombus.
Proof:
Given:
A parallelogram ABCD. Diagonals AC and BD intersect at O such that $\text{AC} \perp \text{BD}$, i.e., $\angle \text{AOB} = 90^\circ$.
To Prove:
ABCD is a rhombus (i.e., all four sides are equal, $\text{AB} = \text{BC} = \text{CD} = \text{DA}$).
Proof:
Since ABCD is a parallelogram, its diagonals bisect each other. So, O is the midpoint of AC and BD.
$\text{AO} = \text{CO}$
(Diagonals of a parallelogram bisect each other) ... (1)
$\text{BO} = \text{DO}$
(Diagonals of a parallelogram bisect each other)
We are given that the diagonals are perpendicular, so $\angle \text{AOB} = 90^\circ$. Since $\angle \text{AOB}$ and $\angle \text{COB}$ form a linear pair on line AC, and they are adjacent angles around point O,
$\angle \text{COB} = 180^\circ - \angle \text{AOB} = 180^\circ - 90^\circ = 90^\circ$
Also, $\angle \text{COD} = \angle \text{AOB} = 90^\circ$ (Vertically Opposite Angles) and $\angle \text{DOA} = \angle \text{COB} = 90^\circ$ (Vertically Opposite Angles).
So, all four angles at the intersection O are right angles.
Consider $\triangle \text{AOB}$ and $\triangle \text{COB}$.
- $\text{AO} = \text{CO}$ (From equation (1))
- $\angle \text{AOB} = \angle \text{COB}$ (Both are $90^\circ$)
- $\text{BO} = \text{BO}$ (Common side)
By the Side-Angle-Side (SAS) congruence rule:
$\triangle \text{AOB} \cong \triangle \text{COB}$
(By SAS)
By CPCTC, the corresponding sides are equal:
$\text{AB} = \text{CB}$
(CPCTC)
Since ABCD is a parallelogram, opposite sides are equal ($\text{AB} = \text{CD}$ and $\text{BC} = \text{DA}$).
We have just proven that $\text{AB} = \text{BC}$. Combining this with the opposite sides property, we get:
$\text{AB} = \text{BC} = \text{CD} = \text{DA}$
Since all four sides of the parallelogram ABCD are equal, it is a rhombus.
Another converse theorem related to the angles bisected by diagonals is also true: If a diagonal of a parallelogram bisects one of the angles, then it is a rhombus.
Properties of a Square
A square is a special type of quadrilateral defined as a parallelogram with all four sides equal in length and one (and therefore all) interior angle measuring $90^\circ$. This definition implies that a square possesses the properties of a parallelogram, a rhombus (all sides equal), and a rectangle (all angles $90^\circ$) simultaneously. Therefore, a square is the most specific type of quadrilateral among parallelograms, rhombuses, and rectangles.
Properties of a Square
Since a square is a parallelogram, a rhombus, and a rectangle, it inherits and combines all the properties of these quadrilaterals. The properties of a square ABCD are:
- Sides: All four sides are equal in length ($\text{AB} = \text{BC} = \text{CD} = \text{DA}$).
- Parallel Sides: Opposite sides are parallel ($\text{AB} \parallel \text{DC}$ and $\text{BC} \parallel \text{DA}$).
- Angles: All four interior angles are right angles, i.e., equal to $90^\circ$ ($\angle \text{A} = \angle \text{B} = \angle \text{C} = \angle \text{D} = 90^\circ$).
- Opposite Angles: Opposite angles are equal (this is covered by all angles being $90^\circ$).
- Consecutive Angles: Consecutive angles are supplementary (this is covered by all angles being $90^\circ$; $90^\circ + 90^\circ = 180^\circ$).
- Diagonals: The diagonals are equal in length ($\text{AC} = \text{BD}$). (Property from Rectangle)
- Diagonals: The diagonals bisect each other (Property from Parallelogram and Rhombus/Rectangle). Let the diagonals intersect at O, so $\text{AO} = \text{CO}$ and $\text{BO} = \text{DO}$.
- Diagonals: The diagonals are perpendicular to each other ($\text{AC} \perp \text{BD}$). (Property from Rhombus)
- Diagonals: The diagonals bisect the angles of the square. Since the angles are $90^\circ$, each diagonal divides the vertex angle into two angles of $45^\circ$ ($90^\circ \div 2 = 45^\circ$). (Property from Rhombus)
Conditions for a Quadrilateral to be a Square
A quadrilateral can be proven to be a square if it satisfies certain conditions. These conditions often build upon the properties of parallelograms, rhombuses, and rectangles.
A quadrilateral is a square if:
- It is a parallelogram with all sides equal and one angle equal to $90^\circ$. (This is the definition)
- It is a rhombus with one angle equal to $90^\circ$. (A rhombus with a right angle becomes a square).
- It is a rhombus with equal diagonals. (A rhombus with equal diagonals must have $90^\circ$ angles, making it a square).
- It is a rectangle with all sides equal. (A rectangle with equal sides becomes a square).
- It is a rectangle with perpendicular diagonals. (A rectangle with perpendicular diagonals must have equal sides, making it a square).
- Its diagonals are equal and bisect each other at right angles. (As proven in the example below).
Example 1. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Answer:
Given:
A quadrilateral ABCD. Its diagonals AC and BD intersect at point O such that $\text{AC} = \text{BD}$, $\text{AO} = \text{CO}$, $\text{BO} = \text{DO}$, and $\text{AC} \perp \text{BD}$ (i.e., $\angle \text{AOB} = 90^\circ$).
To Prove:
Quadrilateral ABCD is a square.
Proof:
We can prove that ABCD is a square by showing it is first a parallelogram, then a rhombus, and then a rectangle.
Step 1: Show that ABCD is a parallelogram.
We are given that the diagonals AC and BD bisect each other at O ($\text{AO} = \text{CO}$ and $\text{BO} = \text{DO}$).
According to Theorem 8.7, if the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
Therefore, ABCD is a parallelogram.
Step 2: Show that ABCD is a rhombus.
We have established that ABCD is a parallelogram. We are also given that the diagonals are perpendicular ($\text{AC} \perp \text{BD}$, so $\angle \text{AOB} = 90^\circ$).
According to Theorem 8.13, if the diagonals of a parallelogram are perpendicular, then it is a rhombus.
Therefore, parallelogram ABCD is a rhombus.
As a rhombus, all its sides are equal:
$\text{AB} = \text{BC} = \text{CD} = \text{DA}$
Step 3: Show that ABCD is a rectangle.
We know that ABCD is a parallelogram. We are given that the diagonals are equal ($\text{AC} = \text{BD}$).
According to Theorem 8.11, if the diagonals of a parallelogram are equal, then it is a rectangle.
Therefore, parallelogram ABCD is a rectangle.
As a rectangle, all its angles are right angles:
$\angle \text{A} = \angle \text{B} = \angle \text{C} = \angle \text{D} = 90^\circ$
Step 4: Conclusion.
We have shown that quadrilateral ABCD is a parallelogram, a rhombus (all sides equal), and a rectangle (all angles $90^\circ$). A quadrilateral that is both a rhombus and a rectangle is a square.
Hence, ABCD is a square.
Thus, if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Mid-Point Theorem
The Mid-Point Theorem is an important result in geometry concerning the line segment formed by joining the midpoints of two sides of a triangle. It establishes a relationship between this line segment and the third side of the triangle.
Theorem 8.9: The Mid-Point Theorem
Theorem 8.9. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
Proof:
Given:
A triangle $\triangle \text{ABC}$. E is the midpoint of side AB, and F is the midpoint of side AC.
To Prove:
(i) $\text{EF} \parallel \text{BC}$
(ii) $\text{EF} = \frac{1}{2} \text{BC}$
Construction:
Extend the line segment EF to a point D such that EF = FD. Join C to D.
Proof:
Consider the triangles $\triangle \text{AEF}$ and $\triangle \text{CDF}$.
We are given that F is the midpoint of AC:
$\text{AF} = \text{CF}$
(Given F is the midpoint of AC) ... (1)
By construction, we extended EF such that EF = FD:
$\text{EF} = \text{FD}$
(By construction) ... (2)
The angles $\angle \text{AFE}$ and $\angle \text{CFD}$ are vertically opposite angles formed by the intersection of lines AC and ED:
$\angle \text{AFE} = \angle \text{CFD}$
(Vertically Opposite Angles) ... (3)
Now, compare $\triangle \text{AEF}$ and $\triangle \text{CDF}$:
- $\text{AF} = \text{CF}$ (From equation (1))
- $\angle \text{AFE} = \angle \text{CFD}$ (From equation (3))
- $\text{EF} = \text{FD}$ (From equation (2))
By the Side-Angle-Side (SAS) congruence rule:
$\triangle \text{AEF} \cong \triangle \text{CDF}$
(By SAS, using (1), (3), (2)) ... (4)
By CPCTC, their corresponding parts are equal:
$\text{AE} = \text{CD}$
(CPCTC) ... (5)
$\angle \text{FAE} = \angle \text{FCD}$
(CPCTC) ... (6)
(Note that $\angle \text{FAE}$ is $\angle \text{BAC}$ or $\angle \text{CAB}$, and $\angle \text{FCD}$ is $\angle \text{ACD}$ or $\angle \text{DCA}$)
In equation (6), $\angle \text{FAE}$ and $\angle \text{FCD}$ are alternate interior angles formed by transversal AC intersecting lines AB (or AE) and CD. Since these alternate interior angles are equal, it implies that AB $\parallel$ CD.
AB $\parallel$ CD
We are given that E is the midpoint of AB, so $\text{AE} = \text{EB}$.
$\text{AE} = \text{EB}$
(Given E is the midpoint of AB) ... (7)
From equations (5) and (7), we have $\text{EB} = \text{CD}$.
$\text{EB} = \text{CD}$
... (8)
Since AB $\parallel$ CD, we also have EB $\parallel$ CD (as E is on AB).
EB $\parallel$ CD
... (9)
Consider quadrilateral EBCD. From equations (8) and (9), one pair of opposite sides (EB and CD) is equal and parallel.
According to Theorem 8.8, if a quadrilateral has a pair of opposite sides that are equal and parallel, then it is a parallelogram.
Therefore, quadrilateral EBCD is a parallelogram.
Now that we know EBCD is a parallelogram, we can use its properties:
Opposite sides of a parallelogram are parallel and equal.
Thus, ED $\parallel$ BC and ED = BC.
For part (i), since ED $\parallel$ BC and EF is a part of the line segment ED, we can conclude that EF $\parallel$ BC.
$\text{EF} \parallel \text{BC}$
(Proved part (i))
For part (ii), we know ED = BC. From our construction, ED is formed by extending EF such that EF = FD. Therefore, ED = EF + FD. Since EF = FD,
$\text{ED} = \text{EF} + \text{EF} = 2\text{EF}$
Substituting ED = BC into this equation:
$\text{BC} = 2\text{EF}$
Dividing both sides by 2:
$\text{EF} = \frac{1}{2} \text{BC}$
(Proved part (ii))
Thus, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it. This completes the proof of the Mid-Point Theorem.
Theorem 8.10: Converse of the Mid-Point Theorem
Theorem 8.10. The line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.
Proof:
Given:
A triangle $\triangle \text{ABC}$. E is the midpoint of side AB (so $\text{AE} = \text{EB}$). A line is drawn through E parallel to side BC, intersecting side AC at point F. Thus, $\text{EF} \parallel \text{BC}$.
To Prove:
F is the midpoint of AC (i.e., $\text{AF} = \text{CF}$).
Construction:
Through point C, draw a line segment CD parallel to AB, intersecting the line EF extended at point D.
Proof:
Consider quadrilateral EBCD.
We are given that $\text{EF} \parallel \text{BC}$. Since the line segment ED contains EF, we have ED $\parallel$ BC.
ED $\parallel$ BC
(Given EF $\parallel$ BC) ... (1)
By construction, we drew CD $\parallel$ AB. Since E lies on AB, we have CD $\parallel$ EB.
CD $\parallel$ EB
(By construction CD $\parallel$ AB) ... (2)
Since both pairs of opposite sides are parallel (from equations (1) and (2)), quadrilateral EBCD is a parallelogram.
Therefore, EBCD is a parallelogram.
In a parallelogram, opposite sides are equal.
$\text{EB} = \text{CD}$
(Opposite sides of parallelogram EBCD) ... (3)
We are given that E is the midpoint of AB, so $\text{AE} = \text{EB}$.
$\text{AE} = \text{EB}$
(Given E is the midpoint of AB) ... (4)
From equations (3) and (4), we can conclude that $\text{AE} = \text{CD}$.
$\text{AE} = \text{CD}$
... (5)
Now, consider $\triangle \text{AEF}$ and $\triangle \text{CDF}$.
From equation (2), CD $\parallel$ AB. Considering transversal ED, the alternate interior angles are equal:
$\angle \text{AEF} = \angle \text{CDF}$
(Alternate Interior Angles, CD $\parallel$ AB, transversal ED) ... (6)
From equation (2), CD $\parallel$ AB. Considering transversal AC, the alternate interior angles are equal:
$\angle \text{EAF} = \angle \text{DCF}$
(Alternate Interior Angles, CD $\parallel$ AB, transversal AC) ... (7)
(Note that $\angle \text{EAF}$ is $\angle \text{BAC}$ or $\angle \text{CAB}$, and $\angle \text{DCF}$ is $\angle \text{ACD}$ or $\angle \text{DCA}$)
Now, compare $\triangle \text{AEF}$ and $\triangle \text{CDF}$:
- $\angle \text{EAF} = \angle \text{DCF}$ (From equation (7))
- $\angle \text{AEF} = \angle \text{CDF}$ (From equation (6))
- $\text{AE} = \text{CD}$ (From equation (5))
By the Angle-Angle-Side (AAS) congruence rule, if two angles and a non-included side of one triangle are equal to the corresponding two angles and the non-included side of another triangle, then the triangles are congruent.
$\triangle \text{AEF} \cong \triangle \text{CDF}$
(By AAS, using (7), (6), (5))
By CPCTC, their corresponding parts are equal:
$\text{AF} = \text{CF}$
(CPCTC)
Since $\text{AF} = \text{CF}$, F is the midpoint of AC.
Thus, F is the midpoint of AC.
This proves that the line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side. This completes the proof of the Converse of the Mid-Point Theorem.
Example 1. In $\triangle \text{ABC}$, E and F are the midpoints of AB and AC respectively. If BC = 8 cm, find the length of EF.
Answer:
Given:
In $\triangle \text{ABC}$, E is the midpoint of side AB, and F is the midpoint of side AC. Length of BC = 8 cm.
To Find:
The length of EF.
Solution:
In $\triangle \text{ABC}$, we are given that E and F are the midpoints of sides AB and AC, respectively.
According to the Mid-Point Theorem (Theorem 8.9), the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
Applying the Mid-Point Theorem to $\triangle \text{ABC}$ and the segment EF:
$\text{EF} = \frac{1}{2} \text{BC}$
We are given the length of BC:
$\text{BC} = 8 \text{ cm}$
(Given)
Substitute the value of BC into the formula from the Mid-Point Theorem:
$\text{EF} = \frac{1}{2} \times 8 \text{ cm}$
$\text{EF} = \frac{8}{2} \text{ cm}$
$\text{EF} = 4 \text{ cm}$
The length of the line segment EF is 4 cm.