Understanding Theorems About Parallelogram and Analyzing Diagonals of a Rectangle and a Rhombus

It may not be understood at first glance, but all squares, rectangles, and rhombi are parallelograms. Therefore, all the properties of parallelograms apply to these quadrilaterals as well! In this lesson, theorems about parallelograms will be discussed to understand this concept better.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Parallelograms
Rectangles
Squares
Rhombi
Triangles and congruent triangles
Rotation

Explore

Investigating Properties of Parallelograms

In the applet below, a parallelogram, a rectangle, a rhombus, and a square can be selected and rotated

9 0^{\circ}

clockwise about the point of intersection of its diagonals. What can be noted when these polygons are rotated

18 0^{\circ} ?

Note that all the quadrilaterals in the applet are parallelograms. Additionally, when rotated

18 0^{\circ},

each quadrilateral is mapped onto itself. Considering this information, what conclusions can be made about the opposite sides of a parallelogram?

Discussion

Properties of a Parallelogram's Opposite Sides

A conclusion that can be made from the previous exploration is that the opposite sides of a parallelogram are congruent. This is explained in detail in the following theorem.

Rule

Parallelogram Opposite Sides Theorem

The opposite sides of a parallelogram are congruent.

In respects to the characteristics of the diagram, the following statement holds true.

$\overline{P Q} ≅ \overline{S R} and \overline{Q R} ≅ \overline{P S}$

Proof

Using Congruent Triangles

This theorem can also be proven by using congruent triangles. Consider the parallelogram $P Q R S$ and its diagonal $\overline{P R} .$

It can be noted that two triangles are formed with

\overline{P R}

as a common side.

△ P Q R and △ R S P

By the definition of a parallelogram,

\overline{P Q}

and

\overline{S R}

are parallel. Therefore, by the Alternate Interior Angles Theorem, it can be stated that

∠ Q P R ≅ ∠ S R P

and that

∠ Q R P ≅ ∠ S P R .

Furthermore, by the Reflexive Property of Congruence,

\overline{P R}

is congruent to itself.

Consequently,

△ P Q R

and

△ R S P

have two pairs of congruent angles and an included congruent side.

∠ Q P R \overline{P R} ∠ Q R P ≅ ∠ S R P ≅ \overline{P R} ≅ ∠ S P R

Therefore, by the Angle-Side-Angle Congruence Theorem,

△ P Q R

and

△ R S P

are congruent triangles.

△ P Q R ≅ △ R S P

Since corresponding parts of congruent figures are congruent,

\overline{P S}

is congruent to

\overline{Q R}

and

\overline{P Q}

is congruent to

\overline{R S} .

$\overline{P Q} ≅ \overline{S R} and \overline{Q R} ≅ \overline{P S}$

Furthermore, it can be stated whether a quadrilateral is a parallelogram just by checking if its opposite sides are congruent.

Rule

Converse Parallelogram Opposite Sides Theorem

If the opposite sides of a quadrilateral are congruent, then the polygon is a parallelogram.

Following the above diagram, the statement below holds true.

If $\overline{P Q} ≅ \overline{S R}$ and $\overline{Q R} ≅ \overline{P S},$ then $P Q R S$ is a parallelogram.

Proof

This theorem can be proven by using congruent triangles. Consider the quadrilateral $P Q R S,$ whose opposite sides are congruent, and its diagonal $\overline{P R} .$ By the Reflexive Property of Congruence, this diagonal is congruent to itself.

Therefore, by the Side-Side-Side Congruence Theorem,

△ P Q R

and

△ R S P

are congruent triangles.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \overline{P Q} ≅ \overline{R S} \overline{Q R} ≅ \overline{S P} \overline{P R} ≅ \overline{P R} \Rightarrow △ P Q R ≅ △ R S P

Since corresponding parts of congruent figures are congruent, corresponding angles of

△ P Q R

and

△ R S P

are congruent.

Finally, by the Converse of the Alternate Interior Angles Theorem, $\overline{P Q}$ is parallel to $\overline{R S}$ and $\overline{Q R}$ is parallel to $\overline{S P} .$ Therefore, by the definition of a parallelogram, $P Q R S$ is a parallelogram.

This proves the theorem.

If $\overline{P Q} ≅ \overline{S R}$ and $\overline{Q R} ≅ \overline{P S},$ then $P Q R S$ is a parallelogram.

Another conclusion that can be made from the exploration is that the opposite angles of a parallelogram are congruent. This is explained in detail in the following theorem.

Rule

Parallelogram Opposite Angles Theorem

In a parallelogram, the opposite angles are congruent.

For the parallelogram $P Q R S,$ the following statement holds true.

$∠ Q ≅ ∠ S and ∠ P ≅ ∠ R$

Proof

This theorem can be proved by using congruent triangles. Consider the parallelogram $P Q R S$ and its diagonal $\overline{P R} .$

Opposite sides of a parallelogram are parallel. Therefore, by the Alternate Interior Angles Theorem it can be stated that $∠ Q P R ≅ ∠ S R P$ and $∠ Q R P ≅ ∠ S P R .$ Furthermore, by the Reflexive Property of Congruence, $\overline{P R}$ is congruent to itself.

Two angles of

△ P Q R

and their included side are congruent to two angles of

△ R S P

and their included side. By the Angle-Side-Angle Congruence Theorem,

△ P Q R

and

△ R S P

are congruent triangles.

△ P Q R ≅ △ R S P

Since corresponding parts of congruent figures are congruent,

∠ Q

and

∠ S

are congruent angles.

By drawing the diagonal $\overline{Q S}$ and using a similar procedure, it can be shown that $∠ P$ and $∠ R$ are also congruent angles.

$∠ Q ≅ ∠ S and ∠ P ≅ ∠ R$

Furthermore, it can be determined whether a quadrilateral is a parallelogram just by looking at its opposite angles.

Rule

Converse Parallelogram Opposite Angles Theorem

If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

A parallelogram with congruent opposite angles

Based on the above diagram, the following statement holds true.

$If$ $∠ A ≅ ∠ C$ $and$ $∠ B ≅ ∠ D,$ $then$ $A B C D$ $is$ $a$ $parallelogram .$

Proof

Assume that $A B C D$ is a quadrilateral with opposite congruent angles. It should be noted that congruent angles have the same measure. Then, let $x^{\circ}$ be the measure of $∠ A$ and $∠ C,$ and $y^{\circ}$ be the measure of $∠ B$ and $∠ D .$

By the Polygon Interior Angles Theorem, the sum of the interior angles of a quadrilateral is

36 0^{\circ} .

With this information, a relation can be found between the consecutive interior angles of

A B C D .

x + y + x + y = 360

▼

Simplify

AddTerms

Add terms

2 x + 2 y = 360

FactorOut

Factor out $2$

2 (x + y) = 360

DivEqn

$LHS / 2 = RHS / 2$

x + y = 180

Since

x + y = 180,

the consecutive interior angles of

A B C D

are supplementary. Therefore, by the Converse Consecutive Interior Angles Theorem, it can be concluded that opposite sides of

A B C D

are parallel.

Consequently, by the definition of a parallelogram,

A B C D

is a parallelogram.

Completed Proof

Given : Prove : ∠ A ≅ ∠ C and ∠ B ≅ ∠ D A B C D is a parallelogram .

Proof:

Example

Solving Problems Using Properties of a Parallelogram

To be able to be carefree and enjoy a soccer match over the weekend, Vincenzo wants to complete his Geometry homework immediately after school. He is given a diagram showing a parallelogram, and asked to find the values of $a,$ $b,$ and $x .$

A parallelogram with its horizontal sides labeled 2x+5 and 5x-10. The four interior angles, starting from the top left corner and proceeding clockwise, are labeled 5a+10, 3b-10, a-10, and 10b+60.

Find the values of

a,

b,

and

x

to help Vincenzo be carefree for the match!

Hint

In a parallelogram, opposite sides are congruent and opposite angles are congruent.

Solution

First, for simplicity, the value of $x$ will be found. After that, the values of $a$ and $b$ will be calculated.

Value of $x$

According to the Parallelogram Opposite Sides Theorem, the opposite sides of a parallelogram are congruent. Therefore, the opposite sides have the same length. With this information, an equation in terms of

x

can be expressed.

2 x + 5 = 5 x - 10

This equation can now be solved for

x .

2 x + 5 = 5 x - 10

▼

Solve for

x

SubEqn

$LHS - 5 x = RHS - 5 x$

- 3 x + 5 = - 10

SubEqn

$LHS - 5 = RHS - 5$

- 3 x = - 15

DivEqn

$LHS / (- 3) = RHS / (- 3)$

x = \frac{- 1 5}{- 3}

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

x = \frac{1 5}{3}

CalcQuot

Calculate quotient

x = 5

Values of $a$ and $b$

According to the Parallelogram Opposite Angles Theorem, the opposite angles of a parallelogram are congruent. That means the opposite angles have the same measure. Knowing this, a system of equations can be expressed.

{(5 a + 10)^{\circ} = (10 b + 60)^{\circ} (a - 10)^{\circ} = (3 b - 10)^{\circ} (I) (II)

This system will be solved by using the Substitution Method. For simplicity, the degree symbol will be removed.

{5 a + 10 = 10 b + 60 a - 10 = 3 b - 10

▼

Solve by substitution

AddEqn

$(II):$ $LHS + 10 = RHS + 10$

{5 a + 10 = 10 b + 60 a = 3 b

Substitute

$(I):$ $a = 3 b$

{5 (3 b) + 10 = 10 b + 60 a = 3 b

Multiply

$(I):$ Multiply

{15 b + 10 = 10 b + 60 a = 3 b

SubEqn

$(I):$ $LHS - 10 = RHS - 10$

{15 b = 10 b + 50 a = 3 b

SubEqn

$(I):$ $LHS - 10 b = RHS - 10 b$

{5 b = 50 a = 3 b

DivEqn

$(I):$ $LHS / 5 = RHS / 5$

{b = 10 a = 3 b

Substitute

$(II):$ $b = 10$

{b = 10 a = 3 (10)

Multiply

$(II):$ Multiply

{b = 10 a = 30

Explore

Investigating the Diagonals of a Parallelogram

In the following applet, a parallelogram is shown. By dragging one of its vertices, different types of parallelograms such as squares, rectangles, and rhombi can be formed. By using the measuring tool provided, investigate what relationships exist between the diagonals of each parallelogram.

After investigating each parallelogram type, what relationships between the diagonals of the parallelograms were discovered?

Discussion

Properties of a Parallelogram's Diagonals

A conclusion that can be made from the previous exploration is that the diagonals of a parallelogram intersect at their midpoint. This is explained in detail in the following theorem.

Rule

Parallelogram Diagonals Theorem

In a parallelogram, the diagonals bisect each other.

If $P Q R S$ is a parallelogram, then the following statement holds true.

$\overline{P M} ≅ \overline{R M} and \overline{Q M} ≅ \overline{S M}$

Proof

Using Congruent Triangles

This theorem can be proven by using congruent triangles. Consider the parallelogram $P Q R S$ and its diagonals $\overline{P R}$ and $\overline{Q S} .$ Let $M$ be the point intersection of the diagonals.

Since $\overline{P Q}$ and $\overline{S R}$ are parallel, by the Alternate Interior Angles Theorem it can be stated that $∠ Q P R ≅ ∠ S R P$ and that $∠ P Q S ≅ ∠ R S Q .$ Furthermore, by the Parallelogram Opposite Sides Theorem it can be said that $\overline{P Q} ≅ \overline{S R} .$

Here, two angles of

△ P M Q

and their included side are congruent to two angles of

△ R M S

and their included side. Therefore, by the Angle-Side-Angle Congruence Theorem

△ P M Q

and

△ R M S

are congruent triangles.

△ P M Q ≅ △ R M S

Since corresponding parts of congruent triangles are congruent,

\overline{P M}

is congruent to

\overline{R M}

and

\overline{Q M}

is congruent to

\overline{S M} .

$\overline{P M} ≅ \overline{R M} and \overline{Q M} ≅ \overline{S M}$

By the definition of a segment bisector, both segments $\overline{P R}$ and $\overline{Q S}$ are bisected at point $M .$ Therefore, it has been proven that the diagonals of a parallelogram bisect each other.

Also, a quadrilateral can be identified as a parallelogram just by looking at its diagonals.

Rule

Converse Parallelogram Diagonals Theorem

If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Quadrilateral with diagonals that bisect each other

Based on the diagram above, the following relation holds true.

If $\overline{A C}$ and $\overline{B D}$ bisect each other, then $A B C D$ is a parallelogram.

Proof

Let $E$ be point of intersection of the diagonals of a quadrilateral. Since the diagonals bisect each other, $E$ is the midpoint of each diagonal.

Because $∠ A E B$ and $∠ C E D$ are vertical angles, they are congruent by the Vertical Angles Theorem. Therefore, by the Side-Angle-Side Congruence Theorem, $△ A E B$ and $△ C E D$ are congruent triangles. Since corresponding parts of congruent figures are congruent, $\overline{A B}$ and $\overline{C D}$ are congruent.

Applying a similar reasoning, it can be concluded that $△ A E D$ and $△ C E B$ are congruent triangles. Consequently, $\overline{A D}$ and $\overline{B C}$ are also congruent.

Finally, since both pairs of opposite sides of quadrilateral $A B C D$ are congruent, the Converse Parallelogram Opposite Sides Theorem states that $A B C D$ is a parallelogram.

Example

Solving Problems Using Properties of a Parallelogram's Diagonals

Vincenzo has one last exercise to finish before going to a soccer match. He has been given a diagram showing a parallelogram. He is asked to find the value of $x$ and $y .$

A parallelogram with its longer diagonal bisected by two segments labeled 9 and 2x+1, and its shorter diagonal bisected by two segments labeled y+4 and 3y.

Find the values of

x

and

y

and help Vincenzo finish his homework!

Hint

The diagonals of a parallelogram bisect each other.

Solution

According to the Parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other.

With this information, two equations can be written.

(I) 9 = 2 x + 1 (II) y + 4 = 3 y

These can be solved one at a time. Equation (I) will be solved first.

9 = 2 x + 1

▼

Solve for

x

SubEqn

$LHS - 1 = RHS - 1$

8 = 2 x

DivEqn

$LHS / 2 = RHS / 2$

4 = x

RearrangeEqn

Rearrange equation

x = 4

The value of

x

4 .

Finally, Equation (II) will be solved.

y + 4 = 3 y

▼

Solve for

y

SubEqn

$LHS - y = RHS - y$

4 = 2 y

DivEqn

$LHS / 2 = RHS / 2$

2 = y

RearrangeEqn

Rearrange equation

y = 2

The value of

y

2 .

Discussion

Investigating Rotations of Parallelograms

Consider a rigid motion that rotates a parallelogram

18 0^{\circ}

about the point of intersection of its diagonals.

Since a rotation is a rigid motion, the preimage and the image are congruent figures. Furthermore, because corresponding parts of congruent figures are congruent, the three statements below hold true.

Opposite sides of a parallelogram are congruent.
Opposite angles of a parallelogram are congruent.
The diagonals of a parallelogram bisect each other.

Note that the Parallelogram Opposite Sides Theorem, the Parallelogram Opposite Angles Theorem, and the Parallelogram Diagonals Theorem have been proved using a

18 0^{\circ}

rotation about the point of intersection of the diagonals.

Discussion

Diagonals of a Rectangle

It can be determined whether a parallelogram is a rectangle just by looking at its diagonals. Furthermore, if a parallelogram is a rectangle, a statement about its diagonals can be made.

Rule

Rectangle Diagonals Theorem

A parallelogram is a rectangle if and only if its diagonals are congruent.

Based on the diagram, the following relation holds true.

$P Q R S$ is a rectangle $\Leftrightarrow$ $\overline{P R} ≅ \overline{Q S}$

Two proofs will be provided for this theorem. Each proof will consist of two parts.

Part I: If $P Q R S$ is a rectangle, then $\overline{P R} ≅ \overline{Q S} .$
Part II: If $\overline{P R} ≅ \overline{Q S},$ then $P Q R S$ is a rectangle.

Proof

Using Similar Triangles

This proof will use similar triangles to prove the theorem.

Part I: $P Q R S$ Is a Rectangle $\Rightarrow \overline{P R} ≅ \overline{Q S}$

Suppose $P Q R S$ is a rectangle and $\overline{P R}$ and $\overline{Q S}$ are its diagonals. By the Parallelogram Opposite Sides Theorem, the opposite sides of a parallelogram are congruent. Therefore, $\overline{R S}$ and $\overline{Q P}$ are congruent. Additionally, by the Reflexive Property of Congruence, $\overline{S P},$ or $\overline{P S},$ is congruent to itself.

Since the angles of a rectangle are right angles, by the definition of congruent angles,

∠ R S P ≅ ∠ Q P S

. Consequently,

△ R S P

and

△ Q P S

have two pairs of congruent sides and congruent included angles.

\overline{R S} ≅ \overline{Q P} ∠ R S P ≅ ∠ Q P S \overline{S P} ≅ \overline{P S}

Therefore, by the Side-Angle-Side Congruence Theorem, the triangles are congruent.

△ R S P ≅ △ Q P S

Because corresponding parts of congruent triangles are congruent,

\overline{P R}

and

\overline{Q S},

which are the diagonals of

P Q R S,

are congruent.

\overline{P R} ≅ \overline{Q S}

Part II: $\overline{P R} ≅ \overline{Q S} \Rightarrow P Q R S$ Is a Rectangle

Consider the parallelogram $P Q R S$ and its diagonals $\overline{P R}$ and $\overline{Q S}$ such that $\overline{P R} ≅ \overline{Q S} .$

By the Parallelogram Opposite Sides Theorem, $\overline{P Q} ≅ \overline{S R} .$ Additionally, by the Reflexive Property of Congruence, $\overline{P S}$ is congruent to itself.

The sides of

△ Q P S

are congruent to the sides of

△ R S P .

\overline{S Q} ≅ \overline{P R} \overline{S R} ≅ \overline{P Q} \overline{P S} ≅ \overline{S P}

Therefore, by the Side-Side-Side Congruence Theorem,

△ Q P S ≅ △ R S P .

Moreover, since corresponding parts of congruent triangles are congruent,

∠ Q P S

is congruent to

∠ R S P .

∠ Q P S ≅ ∠ R S P ⇕ m ∠ Q P S = m ∠ R S P

Note that

∠ Q P S

and

∠ R S P

are consecutive angles. By the Parallelogram Consecutive Angles Theorem, these angles are supplementary. With this information, it can be concluded that both

∠ Q P S

and

∠ R S P

are right angles.

m ∠ Q P S + m ∠ R S P = 18 0^{\circ} ⇓ m ∠ Q P S = 9 0^{\circ} and m ∠ R S P = 9 0^{\circ}

Additionally, by the Parallelogram Opposite Angles Theorem,

∠ Q P S ≅ ∠ S R Q

and

∠ R S P ≅ P Q R .

Because all of the angles are right angles,

P Q R S

is a rectangle.

Proof

Using Transformations

This proof will use transformations to prove the theorem.

Part I: $P Q R S$ Is a Rectangle $\Rightarrow \overline{P R} ≅ \overline{Q S}$

Consider the rectangle $P Q R S$ and its diagonals $\overline{P R}$ and $\overline{Q S} .$ Let $M$ be the point of intersection of the diagonals.

Let $A$ and $B$ be the midpoints of $\overline{P S}$ and $\overline{R Q} .$ Then, a line through $M$ and the midpoints $A$ and $B$ can be drawn.

Note that

\overline{Q A},

\overline{A R},

\overline{P B},

and

\overline{B S}

are congruent segments. Because congruent segments have the same length, the distance between

Q

and

A

equals the distance between

R

and

A .

Therefore,

Q

is the image of

R

after a reflection across

A B .

Similarly,

P

is the image of

S

after the same reflection.

Since

M

lies on

A B,

a reflection across

A B

maps

M

onto itself.

Reflection Across $A B$
Preimage	Image
$R$	$Q$
$S$	$P$
$M$	$M$

The table shows that the images of the vertices of

△ R S M

are the vertices of

△ Q P M .

Therefore,

△ Q P M

is the image of

△ R S M

after a reflection across

A B .

Since a reflection is a rigid motion, this proves that the triangles are congruent.

Because corresponding parts of congruent figures are congruent,

\overline{Q M} ≅ \overline{R M}

and

\overline{P M} ≅ \overline{S M} .

Additionally, by the Parallelogram Diagonals Theorem, the diagonals of the rectangle bisect each other. Therefore, all four segments are congruent.

Each diagonal of the parallelogram consists of the same two congruent segments. By the Segment Addition Postulate, the diagonals are congruent.

\overline{P R} ≅ \overline{Q S}

Part II: $\overline{P R} ≅ \overline{Q S} \Rightarrow P Q R S$ Is a Rectangle

Consider the parallelogram $P Q R S$ and its diagonals $\overline{P R}$ and $\overline{Q S}$ such that $\overline{P R} ≅ \overline{Q S} .$ By the Parallelogram Diagonals Theorem, the diagonals of a rectangle bisect each other at $M .$

By the Parallelogram Opposite Sides Theorem, $\overline{P Q} ≅ \overline{S R}$ and $\overline{Q R} ≅ \overline{P S} .$

Let $A$ and $B$ be the midpoints of $\overline{P S}$ and $\overline{R Q} .$ Then, a line through $M$ and the midpoints $A$ and $B$ can be drawn.

As shown before,

Q,

P,

and

M

are the respective images of

R,

S,

and

M

after a reflection across

A B .

Therefore, since

△ Q P M

is the image of

△ R S M

after a reflection across

A B,

the triangles are congruent.

Let

C

and

D

be the midpoints of of

\overline{P Q}

and

\overline{R S} .

By following the same reasoning,

△ P M S

is the image of

△ Q M R

after a reflection across

C D .

Therefore, the triangles are congruent.

The parallelogram consists of four triangles in which the opposite triangles are congruent. Therefore, the corresponding angles of these triangles are congruent. Additionally, because all triangles are isosceles, the angles opposite congruent sides are congruent as well.

Each angle of the parallelogram is the sum of the same two congruent angles. Therefore, all angles of the parallelogram are congruent.

∠ P ≅ ∠ Q ≅ ∠ R ≅ ∠ S

Moreover, by the Parallelogram Consecutive Angles Theorem,

∠ P

and

∠ S

are supplementary. With this information, it can be concluded that both angles are right triangles.

∠ P + ∠ S = 18 0^{\circ} ⇓ m ∠ P = 9 0^{\circ} and m ∠ S = 9 0^{\circ}

Because all of the angles are congruent, the angles of the parallelogram are right triangles. Therefore,

P Q R S

is a rectangle.

Example

Solving Problems Using the Diagonals of a Rectangle

Zosia arrives early to a Harry Styles concert! She notices something about the stage, so she uses a napkin as paper and draws a diagram. The stage is a rectangle that she labels as $A B C D .$

A rectangle with vertices labeled A, B, C, and D in clockwise order, starting from the top left corner. Diagonal AC is labeled 4x+3, and diagonal BC is labeled 45-2x.

Find the lengths of its diagonals to help Zosia understand the stage that she will see her favorite artist sing on!

Hint

In a rectangle, the diagonals are congruent.

Solution

By the Rectangle Diagonals Theorem, the diagonals of a rectangle are congruent. This means that

\overline{A C}

and

\overline{B D}

have the same length. With this information, an equation in terms of

x

can be written.

4 x + 3 = 45 - 2 x

The above equation will be solved for

x .

4 x + 3 = 45 - 2 x

▼

Solve for

x

AddEqn

$LHS + 2 x = RHS + 2 x$

6 x + 3 = 45

SubEqn

$LHS - 3 = RHS - 3$

6 x = 42

DivEqn

$LHS / 6 = RHS / 6$

x = 7

The value of

x

was found. Next, this value can be substituted in any of the expressions for the length of a diagonal. In this case,

x = 7

will be arbitrarily substituted in the expression for

A C .

A C = 4 x + 3

Substitute

$x = 7$

A C = 4 (7) + 3

▼

Evaluate right-hand side

Multiply

A C = 28 + 3

AddTerms

Add terms

A C = 31

It was found that the length of

\overline{A C}

31 .

Since the diagonals of a rectangle are congruent, the length of

\overline{B D}

is also

31 .

A C = 31 and B D = 31

Discussion

Diagonals of a Rhombus

As with rectangles, it can also be determined whether a parallelogram is a rhombus just by looking at its diagonals.

Rule

Rhombus Diagonals Theorem

A parallelogram is a rhombus if and only if its diagonals are perpendicular.

Based on the diagram, the following relation holds true.

Parallelogram $A B C D$ is a rhombus $\Leftrightarrow$ $\overline{A C} ⊥ \overline{B D}$

Proof

This proof will be written in two parts.

If a Parallelogram Is a Rhombus, Then Its Diagonals Are Perpendicular

A rhombus is a parallelogram with four congruent sides. By the Parallelogram Diagonals Theorem, it can be said that its diagonals bisect each other. Let Let $A B C D$ be a rhombus with $P$ at the midpoint of both diagonals.

Note that

\overline{A B}

is congruent to

\overline{A D}

and

\overline{D P}

is congruent to

\overline{B P} .

Additionally, by the Reflexive Property of Congruence,

\overline{A P}

is congruent to itself. Therefore, by the Side-Side-Side Congruence Theorem,

△ A P B

is congruent to

△ A P D .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \overline{A B} ≅ \overline{A D} \overline{D P} ≅ \overline{B P} \overline{A P} ≅ \overline{A P} \Rightarrow △ A P B ≅ △ A P D

Because corresponding parts of congruent triangles are congruent,

∠ A P B

and

∠ A P D

are congruent angles. Furthermore, these angles form a linear pair, which means they are supplementary. With this information, it can be concluded that both

∠ A P B

and

∠ A P D

are right angles.

{∠ A P B ≅ ∠ A P D m ∠ A P B + m ∠ A P D = 18 0^{\circ} ⇓ m ∠ A P B = 9 0^{\circ} and m ∠ A P D = 9 0^{\circ}

This implies that

\overline{A C}

is perpendicular to

\overline{B D} .

Therefore, the diagonals of a rhombus are perpendicular.

Parallelogram $A B C D$ is a rhombus $\Rightarrow$ $\overline{A C} ⊥ \overline{B D}$

If Its Diagonals Are Perpendicular, Then a Parallelogram is a Rhombus

Conversely, let $A B C D$ be a parallelogram whose diagonals are perpendicular.

By the Parallelogram Diagonals Theorem, the diagonals of the parallelogram bisect each other. If $P$ is the midpoint of both diagonals, then $\overline{A P}$ and $\overline{C P}$ are congruent.

Since

\overline{A C}

and

\overline{B D}

are perpendicular,

∠ A P B

and

∠ C P B

measure

9 0^{\circ}

and thus are congruent angles. By the Reflexive Property of Congruence,

\overline{B P}

is congruent to itself. This means that two sides and their included angle are congruent. By the Side-Angle-Side Congruence Theorem,

△ A P B

and

△ C P B

are congruent triangles.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \overline{A P} ≅ \overline{C P} ∠ A P B ≅ ∠ C P B \overline{B P} ≅ \overline{B P} ⇓ △ A P B ≅ △ C P B

Because corresponding parts of congruent figures are congruent, it can be said that

\overline{A B}

is congruent to

\overline{C B} .

Furthermore, by the Parallelogram Opposite Sides Theorem, $\overline{A B}$ is congruent to $\overline{D C}$ and $\overline{A D}$ is congruent to $\overline{B C} .$ By the Transitive Property of Congruence, it follows that all sides of the parallelogram are congruent.

This means that if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.

$\overline{A C} ⊥ \overline{B D}$ $\Rightarrow$ parallelogram $A B C D$ is a rhombus

Example

Solving Problems Using the Diagonals of a Rhombus

Zosia is now listening to Dua Lipa at home. Staring at some of her album covers, Zosia decides to design a parallelogram as the background art for Dua's next cover! She has made a parallelogram $A B C D$ in which the diagonals are perpendicular. To make a unique design, she wants to be sure of the length of $\overline{A B} .$

Help Zosia draw the perfect design by finding the length of

\overline{A B}!

Hint

If the diagonals of a parallelogram are perpendicular, then the quadrilateral is a rhombus.

Solution

By the Rhombus Diagonal Theorem, the diagonals of a parallelogram are perpendicular if and only if the quadrilateral is a rhombus. Therefore, since

\overline{B D}

and

\overline{A C}

are perpendicular,

A B C D

is a rhombus. This means that

\overline{B C}

and

\overline{C D}

have the same length. With this information, an equation in terms of

x

can be written.

6 x + 3 = 8 x - 19

The above equation will be now solved for

x .

6 x + 3 = 8 x - 19

▼

SubEqn

$LHS - 8 x = RHS - 8 x$

- 2 x + 3 = - 19

SubEqn

$LHS - 3 = RHS - 3$

- 2 x = - 22

DivEqn

$LHS / (- 2) = RHS / (- 2)$

x = \frac{- 2 2}{- 2}

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

x = \frac{2 2}{2}

CalcQuot

Calculate quotient

x = 11

The value of

x

was found. Since the given quadrilateral is a rhombus, all sides are congruent and therefore have the same length. This means that the length of

\overline{A B}

is the same as the length of

\overline{B C} .

To find it,

x = 11

will be substituted into the expression for

B C .

B C = 6 x + 3

Substitute

$x = 11$

B C = 6 (11) + 3

▼

Evaluate right-hand side

Multiply

B C = 66 + 3

AddTerms

Add terms

B C = 69

The length of

\overline{B C}

69 .

As it has been previously said, all sides have the same length. Therefore,

A B

is also

69 .

Closure

Additional Applications of a Parallelogram's Properties

By using the theorems seen in this lesson, other properties can be derived. One of them is the Parallelogram Consecutive Angles Theorem.

Parallelogram Consecutive Angles Theorem
The consecutive angles of a parallelogram are supplementary.

Furthermore, the theorems seen in this lesson can be applied to different parallelograms in different contexts. Consider a square. By definition, all its angles are right angles, and all its sides are congruent. Therefore, a square is both a rectangle and a rhombus.

Therefore, by the Rectangle Diagonals Theorem and the Rhombus Diagonals Theorem, the diagonals of a square are congruent and perpendicular.

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Catch-Up and Review

Proof

Proof

Proof

Proof

Completed Proof

Hint

Solution

Value of x

Values of a and b

Proof

Proof

Hint

Solution

Proof

Part I: PQRS Is a Rectangle ⇒PR≅QS​

Part II: PR≅QS​⇒PQRS Is a Rectangle

Proof

Part I: PQRS Is a Rectangle ⇒PR≅QS​

Part II: PR≅QS​⇒PQRS Is a Rectangle

Hint

Solution

Proof

If a Parallelogram Is a Rhombus, Then Its Diagonals Are Perpendicular

If Its Diagonals Are Perpendicular, Then a Parallelogram is a Rhombus

Hint

Solution

Value of $x$

Values of $a$ and $b$

Part I: $P Q R S$ Is a Rectangle $\Rightarrow \overline{P R} ≅ \overline{Q S}$

Part II: $\overline{P R} ≅ \overline{Q S} \Rightarrow P Q R S$ Is a Rectangle

Part I: $P Q R S$ Is a Rectangle $\Rightarrow \overline{P R} ≅ \overline{Q S}$

Part II: $\overline{P R} ≅ \overline{Q S} \Rightarrow P Q R S$ Is a Rectangle