13. Theorems About Parallelograms
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Chapter 1
13.
Theorems About Parallelograms
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Why
In geometry, parallelograms hold a significant place due to their unique properties. One of the fascinating aspects of parallelograms is the behavior of their diagonals. Specifically, the diagonals of a rectangle are always congruent, meaning they have the same length. On the other hand, the diagonals of a rhombus intersect each other at right angles, making them perpendicular. These properties are not just theoretical; they have practical applications in various fields, from architecture to design. Understanding these theorems can provide a solid foundation for further exploration in geometry, especially when dealing with quadrilaterals and their characteristics.
Lesson Settings & Tools
| | 14 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Theorems About Parallelograms
Slide of 14
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.
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Explore
Investigating Properties of Parallelograms
In the applet below, a parallelogram, a rectangle, a rhombus, and a square can be selected and rotated 90^(∘) clockwise about the point of intersection of its diagonals. What can be noted when these polygons are rotated 180^(∘)?
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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.
PQ≅SR and QR≅PS
Proof
Using Congruent TrianglesThis theorem can also be proven by using congruent triangles. Consider the parallelogram PQRS and its diagonal PR.
It can be noted that two triangles are formed with PR as a common side. △ PQR and △ RSP By the definition of a parallelogram, PQ and SR are parallel. Therefore, by the Alternate Interior Angles Theorem, it can be stated that ∠ QPR ≅ ∠ SRP and that ∠ QRP ≅ ∠ SPR. Furthermore, by the Reflexive Property of Congruence, PR is congruent to itself.
Consequently, △ PQR and △ RSP have two pairs of congruent angles and an included congruent side. ∠ QPR &≅ ∠ SRP PR&≅PR ∠ QRP &≅ ∠ SPR Therefore, by the Angle-Side-Angle Congruence Theorem, △ PQR and △ RSP are congruent triangles. △ PQR ≅ △ RSP Since corresponding parts of congruent figures are congruent, PS is congruent to QR and PQ is congruent to RS.
PQ≅SR and QR≅PS
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 PQ≅SR and QR≅PS, then PQRS is a parallelogram.
Proof
This theorem can be proven by using congruent triangles. Consider the quadrilateral PQRS, whose opposite sides are congruent, and its diagonal PR. By the Reflexive Property of Congruence, this diagonal is congruent to itself.
Therefore, by the Side-Side-Side Congruence Theorem, △ PQR and △ RSP are congruent triangles. PQ≅ RS QR≅ SP PR≅ PR ⇒ △ PQR ≅ △ RSP Since corresponding parts of congruent figures are congruent, corresponding angles of △ PQR and △ RSP are congruent.
Finally, by the Converse of the Alternate Interior Angles Theorem, PQ is parallel to RS and QR is parallel to SP. Therefore, by the definition of a parallelogram, PQRS is a parallelogram.
This proves the theorem.
If PQ≅SR and QR≅PS, then PQRS is a parallelogram.
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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 PQRS, the following statement holds true.
∠ Q ≅ ∠ S and ∠ P ≅ ∠ R
Proof
This theorem can be proved by using congruent triangles. Consider the parallelogram PQRS and its diagonal PR.
Opposite sides of a parallelogram are parallel. Therefore, by the Alternate Interior Angles Theorem it can be stated that ∠QPR≅∠SRP and ∠QRP≅∠SPR. Furthermore, by the Reflexive Property of Congruence, PR is congruent to itself.
Two angles of △PQR and their included side are congruent to two angles of △RSP and their included side. By the Angle-Side-Angle Congruence Theorem, △PQR and △RSP are congruent triangles. △PQR≅△RSP Since corresponding parts of congruent figures are congruent, ∠Q and ∠S are congruent angles.
By drawing the diagonal QS 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.
Based on the above diagram, the following statement holds true.
If ∠ A ≅ ∠ C and ∠ B ≅ ∠ D, then ABCD is a parallelogram.
Proof
Assume that ABCD is a quadrilateral with opposite congruent angles. It should be noted that congruent angles have the same measure. Then, let x^(∘) be the measure of ∠ A and ∠ C, and y^(∘) be the measure of ∠ B and ∠ D.
By the Polygon Interior Angles Theorem, the sum of the interior angles of a quadrilateral is 360^(∘). With this information, a relation can be found between the consecutive interior angles of ABCD.
x+y+x+y=360
x+y=180
Since x+y=180, the consecutive interior angles of ABCD are supplementary. Therefore, by the Converse Consecutive Interior Angles Theorem, it can be concluded that opposite sides of ABCD are parallel.
Consequently, by the definition of a parallelogram, ABCD is a parallelogram.
Completed Proof
2 &Given:&& ∠ A ≅ ∠ C and ∠ B ≅ ∠ D &Prove:&& ABCD is a parallelogram. Proof:
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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.
Find the values of a, b, and x to help Vincenzo be carefree for the match!
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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. 2x+5=5x-10 This equation can now be solved for x.
2x+5=5x-10
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. (5a+10)^(∘)=(10b+60)^(∘) & (I) (a-10)^(∘)=(3b-10)^(∘) & (II) This system will be solved by using the Substitution Method. For simplicity, the degree symbol will be removed.
5a+10=10b+60 a-10=3b-10
▼
Solve by substitution
AddEqn
(II): LHS+10=RHS+10
5a+10=10b+60 a=3b
Substitute
(I): a= 3b
5( 3b)+10=10b+60 a=3b
Multiply
(I): Multiply
15b+10=10b+60 a=3b
SubEqn
(I): LHS-10=RHS-10
15b=10b+50 a=3b
SubEqn
(I): LHS-10b=RHS-10b
5b=50 a=3b
DivEqn
(I): .LHS /5.=.RHS /5.
b=10 a=3b
Substitute
(II): b= 10
b=10 a=3( 10)
Multiply
(II): Multiply
b=10 a=30
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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.
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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 PQRS is a parallelogram, then the following statement holds true.
PM≅RM and QM≅SM
Proof
Using Congruent TrianglesThis theorem can be proven by using congruent triangles. Consider the parallelogram PQRS and its diagonals PR and QS. Let M be the point intersection of the diagonals.
Since PQ and SR are parallel, by the Alternate Interior Angles Theorem it can be stated that ∠ QPR ≅ ∠ SRP and that ∠ PQS ≅ ∠ RSQ. Furthermore, by the Parallelogram Opposite Sides Theorem it can be said that PQ≅SR.
Here, two angles of △PMQ and their included side are congruent to two angles of △RMS and their included side. Therefore, by the Angle-Side-Angle Congruence Theorem △PMQ and △RMS are congruent triangles. △PMQ≅△RMS Since corresponding parts of congruent triangles are congruent, PM is congruent to RM and QM is congruent to SM.
PM≅RM and QM≅SM
By the definition of a segment bisector, both segments PR and QS 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.
Based on the diagram above, the following relation holds true.
If AC and BD bisect each other, then ABCD 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 ∠ AEB and ∠ CED are vertical angles, they are congruent by the Vertical Angles Theorem. Therefore, by the Side-Angle-Side Congruence Theorem, △ AEB and △ CED are congruent triangles. Since corresponding parts of congruent figures are congruent, AB and CD are congruent.
Applying a similar reasoning, it can be concluded that △ AED and △ CEB are congruent triangles. Consequently, AD and BC are also congruent.
Finally, since both pairs of opposite sides of quadrilateral ABCD are congruent, the Converse Parallelogram Opposite Sides Theorem states that ABCD is a parallelogram.
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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.
Find the values of x and y and help Vincenzo finish his homework!
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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=2x+1 &(II) y+4=3y These can be solved one at a time. Equation (I) will be solved first.
9=2x+1
▼
Solve for x
x=4
The value of x is 4. Finally, Equation (II) will be solved.
y+4=3y
▼
Solve for y
y=2
The value of y is 2.
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Discussion
Investigating Rotations of Parallelograms
Consider a rigid motion that rotates a parallelogram 180^(∘) about the point of intersection of its diagonals.
- Opposite sides of a parallelogram are congruent.
- Opposite angles of a parallelogram are congruent.
- The diagonals of a parallelogram bisect each other.
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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.
PQRS is a rectangle ⇔ PR ≅ QS
Two proofs will be provided for this theorem. Each proof will consist of two parts.
- Part I: If PQRS is a rectangle, then PR≅QS.
- Part II: If PR≅QS, then PQRS is a rectangle.
Proof
Using Similar TrianglesThis proof will use similar triangles to prove the theorem.
Part I: PQRS Is a Rectangle ⇒ PR≅QS
Suppose PQRS is a rectangle and PR and QS are its diagonals. By the Parallelogram Opposite Sides Theorem, the opposite sides of a parallelogram are congruent. Therefore, RS and QP are congruent. Additionally, by the Reflexive Property of Congruence, SP, or PS, is congruent to itself.
Since the angles of a rectangle are right angles, by the definition of congruent angles, ∠ RSP ≅ ∠ QPS. Consequently, △ RSP and △ QPS have two pairs of congruent sides and congruent included angles. RS≅QP ∠ RSP ≅ ∠ QPS SP≅PS Therefore, by the Side-Angle-Side Congruence Theorem, the triangles are congruent. △ RSP≅△ QPS Because corresponding parts of congruent triangles are congruent, PR and QS, which are the diagonals of PQRS, are congruent. PR≅ QS
Part II: PR≅QS ⇒ PQRS Is a Rectangle
Consider the parallelogram PQRS and its diagonals PR and QS such that PR≅QS.
By the Parallelogram Opposite Sides Theorem, PQ≅SR. Additionally, by the Reflexive Property of Congruence, PS is congruent to itself.
The sides of △ QPS are congruent to the sides of △ RSP. SQ≅PR SR≅PQ PS≅SP Therefore, by the Side-Side-Side Congruence Theorem, △ QPS≅ △ RSP. Moreover, since corresponding parts of congruent triangles are congruent, ∠ QPS is congruent to ∠ RSP. ∠ QPS ≅ ∠ RSP ⇕ m∠ QPS = m∠ RSP Note that ∠ QPS and ∠ RSP are consecutive angles. By the Parallelogram Consecutive Angles Theorem, these angles are supplementary. With this information, it can be concluded that both ∠ QPS and ∠ RSP are right angles. m∠ QPS + m∠ RSP = 180^(∘) ⇓ m∠ QPS=90^(∘) and m∠ RSP=90^(∘) Additionally, by the Parallelogram Opposite Angles Theorem, ∠ QPS ≅ ∠ SRQ and ∠ RSP≅ PQR. Because all of the angles are right angles, PQRS is a rectangle.
Proof
Using TransformationsThis proof will use transformations to prove the theorem.
Part I: PQRS Is a Rectangle ⇒ PR≅QS
Consider the rectangle PQRS and its diagonals PR and QS. Let M be the point of intersection of the diagonals.
Let A and B be the midpoints of PS and RQ. Then, a line through M and the midpoints A and B can be drawn.
Note that QA, AR, PB, and BS 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 AB. Similarly, P is the image of S after the same reflection.
Since M lies on AB, a reflection across AB maps M onto itself.
| Reflection Across AB | |
|---|---|
| Preimage | Image |
| R | Q |
| S | P |
| M | M |
The table shows that the images of the vertices of △ RSM are the vertices of △ QPM. Therefore, △ QPM is the image of △ RSM after a reflection across AB. Since a reflection is a rigid motion, this proves that the triangles are congruent.
Because corresponding parts of congruent figures are congruent, QM≅RM and PM≅SM. 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. PR≅QS
Part II: PR≅QS ⇒ PQRS Is a Rectangle
Consider the parallelogram PQRS and its diagonals PR and QS such that PR≅QS. By the Parallelogram Diagonals Theorem, the diagonals of a rectangle bisect each other at M.
By the Parallelogram Opposite Sides Theorem, PQ≅SR and QR≅PS.
Let A and B be the midpoints of PS and RQ. 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 AB. Therefore, since △QPM is the image of △RSM after a reflection across AB, the triangles are congruent.
Let C and D be the midpoints of of PQ and RS. By following the same reasoning, △PMS is the image of △QMR after a reflection across CD. 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=180^(∘) ⇓ m∠P=90^(∘) and m∠S=90^(∘) Because all of the angles are congruent, the angles of the parallelogram are right triangles. Therefore, PQRS is a rectangle.
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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 ABCD.
Find the lengths of its diagonals to help Zosia understand the stage that she will see her favorite artist sing on!
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Hint
In a rectangle, the diagonals are congruent.
Solution
By the Rectangle Diagonals Theorem, the diagonals of a rectangle are congruent. This means that AC and BD have the same length. With this information, an equation in terms of x can be written.
4x+3=45-2x
The above equation will be solved for x.
4x+3=45-2x
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 AC.
It was found that the length of AC is 31. Since the diagonals of a rectangle are congruent, the length of BD is also 31. AC=31 and BD=31
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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 ABCD is a rhombus ⇔ AC⊥BD
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 ABCD be a rhombus with P at the midpoint of both diagonals.
Note that AB is congruent to AD and DP is congruent to BP. Additionally, by the Reflexive Property of Congruence, AP is congruent to itself. Therefore, by the Side-Side-Side Congruence Theorem, △ APB is congruent to △ APD. AB≅ AD DP≅ BP AP≅ AP ⇒ △ APB ≅ △ APD Because corresponding parts of congruent triangles are congruent, ∠ APB and ∠ APD are congruent angles. Furthermore, these angles form a linear pair, which means they are supplementary. With this information, it can be concluded that both ∠ APB and ∠ APD are right angles. ∠ APB ≅ ∠ APD m∠ APB + m∠ APD = 180 ^(∘) ⇓ m∠ APB = 90^(∘) and m∠ APD = 90^(∘) This implies that AC is perpendicular to BD. Therefore, the diagonals of a rhombus are perpendicular.
Parallelogram ABCD is a rhombus ⇒ AC⊥BD
If Its Diagonals Are Perpendicular, Then a Parallelogram is a Rhombus
Conversely, let ABCD 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 AP and CP are congruent.
Since AC and BD are perpendicular, ∠ APB and ∠ CPB measure 90^(∘) and thus are congruent angles. By the Reflexive Property of Congruence, BP is congruent to itself. This means that two sides and their included angle are congruent. By the Side-Angle-Side Congruence Theorem, △ APB and △ CPB are congruent triangles. AP≅ CP ∠ APB ≅ ∠ CPB BP≅BP ⇓ △APB≅ △CPB Because corresponding parts of congruent figures are congruent, it can be said that AB is congruent to CB.
Furthermore, by the Parallelogram Opposite Sides Theorem, AB is congruent to DC and AD is congruent to BC. 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.
AC⊥BD ⇒ parallelogram ABCD is a rhombus
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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 ABCD in which the diagonals are perpendicular. To make a unique design, she wants to be sure of the length of AB.
Help Zosia draw the perfect design by finding the length of AB!
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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 BD and AC are perpendicular, ABCD is a rhombus. This means that BC and CD have the same length. With this information, an equation in terms of x can be written.
6x+3=8x-19
The above equation will be now solved for x.
6x+3=8x-19
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 AB is the same as the length of BC. To find it, x=11 will be substituted into the expression for BC.
The length of BC is 69. As it has been previously said, all sides have the same length. Therefore, AB is also 69.
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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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Theorems About Parallelograms
Exercise 3.1
In ABDC, we know that m∠ BAD=40^(∘), m∠ CAD=x^2, and m∠ BDC=13x.
Determine m∠ CAD if ∠ BDC is an acute angle.
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We are given some properties of the parallelogram. Let's add them to the diagram.
According to the Parallelogram Opposite Angles Theorem, opposite angles of a parallelogram are congruent. Therefore, we can set the opposite corners of ABDC equal to each other in an equation. ccccc m∠ BAD -5pt & + & -5pt m∠ CAD -5pt & = & -5pt m∠ BDC 40 -5pt & + & -5pt x^2 -5pt & = & -5pt 13x Let's rewrite the equation in standard form.
Now we can use the Quadratic Formula to solve for x.
We found two possible solutions. Recall that we were given a restriction on ∠ BDC. It must be an acute angle which means its measure must be less than 90^(∘). Let's calculate the value of m∠ BDC when x_1= 8 and x_2= 5 to see which meets the restriction. 13 x_1=13( 8)&=104^(∘) (obtuse) 13 x_2=13( 5)&=64^(∘) (acute) As we can see only x= 5 results in an acute angle. Therefore, let's use this value to evaluate the measure of ∠ CAD. m∠ CAD= x^2 ⇓ m∠ CAD= 5^2=25^(∘) The measure of ∠ CAD is 25^(∘).
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E
Hint & Answer
S
Solution
What is the measure of ∠ EPF?
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Examining the diagram we notice that EJ and FI are both angle bisectors of ∠ E and ∠ F respectively. This means that the segments cut the angles in two congruent halves. Let's call the measures of these angles x and y, respectively.
In a parallelogram, any two consecutive angles are supplementary angles. Since ∠ CEF and ∠ EFD are consecutive, we can write the following equation. 2x+2y=180^(∘) ⇓ x+y=90^(∘) Now that we have found the sum of x and y, we turn our attention to the triangle EPF.
Using the Interior Angles Theorem, we can write an equation. m∠ EPF+(x+y)=180^(∘) Since we know the measure of (x+y), we can determine the measure of ∠ EPF.
As we can see, ∠ EPF is a right angle.
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Theorems About Parallelograms
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