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Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Prove a quadrilateral is a parallelogram.
• Show a quadrilateral is a parallelogram in the $x-y$ plane.

Review Queue

1. Plot the points $A(2, 2), B(4, -2), C(-2, -4)$, and $D(-6, -2)$.
1. Find the slopes of $\overline{AB},\overline{BC}, \overline{CD},$ and $\overline{AD}$. Is $ABCD$ a parallelogram?
2. Find the point of intersection of the diagonals by finding the midpoint of each.

Know What? You are marking out a baseball diamond and standing at home plate. $3^{rd}$ base is 90 feet away, $2^{nd}$ base is 127.3 feet away, and $1^{st}$ base is also 90 feet away. The angle at home plate is $90^\circ$, from $1^{st}$ to $3^{rd}$ is $90^\circ$. Find the length of the other diagonal (using the Pythagorean Theorem) and determine if the baseball diamond is a parallelogram.

Determining if a Quadrilateral is a Parallelogram

The converses of the theorems in the last section will now be used to see if a quadrilateral is a parallelogram.

Opposite Sides Theorem Converse: If the opposite sides of a quadrilateral are congruent, then the figure is a parallelogram.

If then

Opposite Angles Theorem Converse: If the opposite angles of a quadrilateral are congruent, then the figure is a parallelogram.

If then

Parallelogram Diagonals Theorem Converse: If the diagonals of a quadrilateral bisect each other, then the figure is a parallelogram.

If then

Proof of the Opposite Sides Theorem Converse

Given: $\overline{AB} \cong \overline{DC}, \overline{AD} \cong \overline{BC}$

Prove: $ABCD$ is a parallelogram

Statement Reason
1. $\overline{AB} \cong \overline{DC}, \overline{AD} \cong \overline{BC}$ Given
2. $\overline{DB} \cong \overline{DB}$ Reflexive PoC
3. $\triangle ABD \cong \triangle CDB$ SSS
4. $\angle ABD \cong \angle BDC, \angle ADB \cong \angle DBC$ CPCTC
5. $\overline{AB} \| \overline{DC}, \overline{AD} \| \overline{BC}$ Alternate Interior Angles Converse
6. $ABCD$ is a parallelogram Definition of a parallelogram

Example 1: Write a two-column proof.

Given: $\overline{AB} \| \overline{DC}$, and $\overline{AB} \cong \overline{DC}$

Prove: $ABCD$ is a parallelogram

Solution:

Statement Reason
1. $\overline{AB} \| \overline{DC}$, and $\overline{AB} \cong \overline{DC}$ Given
2. $\angle ABD \cong \angle BDC$ Alternate Interior Angles
3. $\overline{DB} \cong \overline{DB}$ Reflexive PoC
4. $\triangle ABD \cong \triangle CDB$ SAS
5. $\overline{AD} \cong \overline{BC}$ CPCTC
6. $ABCD$ is a parallelogram Opposite Sides Converse

Theorem 6-10: If a quadrilateral has one set of parallel lines that are also congruent, then it is a parallelogram.

If then

Example 2: Is quadrilateral $EFGH$ a parallelogram? How do you know?

Solution:

a) By the Opposite Angles Theorem Converse, $EFGH$ is a parallelogram.

b) $EFGH$ is not a parallelogram because the diagonals do not bisect each other.

Example 3: Algebra Connection What value of $x$ would make $ABCD$ a parallelogram?

Solution: $\overline{AB} \| \overline{DC}$. By Theorem 6-10, $ABCD$ would be a parallelogram if $AB = DC$.

$5x - 8 & = 2x + 13\\3x & = 21\\x & = 7$

Showing a Quadrilateral is a Parallelogram in the $x-y$ Plane

To show that a quadrilateral is a parallelogram in the $x-y$ plane, you might need:

• The Slope Formula, $\frac{y_2 - y_1}{x_2 - x_1}$.
• The Distance Formula, $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
• The Midpoint Formula, $\left ( \frac{x_1 + x_2 }{2} , \frac{y_1 + y_2}{2} \right )$.

Example 4: Is the quadrilateral $ABCD$ a parallelogram?

Solution: Let’s use Theorem 6-10 to see if $ABCD$ is a parallelogram. First, find the length of $AB$ and $CD$.

$AB & = \sqrt{(-1-3)^2 + (5 - 3)^2} && CD = \sqrt{(2 - 6)^2 + (-2 + 4)^2}\\& = \sqrt{(-4)^2 + 2^2} && = \sqrt{(-4)^2 + 2^2}\\& = \sqrt{16 + 4} && = \sqrt{16 + 4}\\& = \sqrt{20} &&= \sqrt{20}$

Find the slopes.

$\text{Slope}\ AB = \frac{5 - 3}{-1-3} = \frac{2}{-4} = -\frac{1}{2} \qquad \text{Slope}\ CD = \frac{-2 +4}{2-6} = \frac{2}{-4} = -\frac{1}{2}$

$AB = CD$ and the slopes are the same, $ABCD$ is a parallelogram.

Example 5: Is the quadrilateral $RSTU$ a parallelogram?

Solution: Let’s use the Parallelogram Diagonals Converse to see if $RSTU$ is a parallelogram. Find the midpoint of each diagonal.

$&\text{Midpoint of}\ RT = \left ( \frac{-4 + 3}{2},\frac{3 - 4}{2}\right ) = (-0.5,-0.5)\\&\text{Midpoint of}\ SU = \left ( \frac{4 - 5}{2}, \frac{5 - 5}{2} \right ) = (-0.5,0)$

$RSTU$ is not a parallelogram because the midpoints are not the same.

Know What? Revisited Use the Pythagorean Theorem to find the length of the second diagonal.

$90^2 + 90^2 & = d^2\\8100 + 8100 & = d^2\\16200 & = d^2\\d & = 127.3$

The diagonals are equal, so the other two sides of the diamond must also be 90 feet. The baseball diamond is a parallelogram, and more specifically, a square.

Review Questions

• Questions 1-12 are similar to Example 2.
• Questions 13-15 are similar to Example 3.
• Questions 16-22 are similar to Examples 4 and 5.
• Questions 23-25 are similar to Example 1 and the proof of the Opposite Sides Converse.

For questions 1-12, determine if the quadrilaterals are parallelograms.

Algebra Connection For questions 13-18, determine the value of $x$ and $y$ that would make the quadrilateral a parallelogram.

For questions 19-22, determine if $ABCD$ is a parallelogram.

1. $A(8, -1), B(6, 5), C(-7, 2), D(-5, -4)$
2. $A(-5, 8), B(-2, 9), C(3, 4), D(0, 3)$
3. $A(-2, 6), B(4, -4), C(13, -7), D(4, -10)$
4. $A(-9, -1), B(-7, 5), C(3, 8), D(1, 2)$

Fill in the blanks in the proofs below.

1. Opposite Angles Theorem Converse

Given: $\angle A \cong \angle C, \angle D \cong \angle B$

Prove: $ABCD$ is a parallelogram

Statement Reason
1.
2. $m \angle A = m \angle C, m \angle D = m \angle B$
4. $m \angle A + m \angle A + m \angle B + m \angle B = 360^\circ$
5. Combine Like Terms
6. Division PoE

7. $\angle A$ and $\angle B$ are supplementary

$\angle A$ and $\angle D$ are supplementary

8. Consecutive Interior Angles Converse
9. $ABCD$ is a parallelogram
1. Parallelogram Diagonals Theorem Converse

Given: $\overline{AE} \cong \overline{EC}, \overline{DE} \cong \overline{EB}$

Prove: $ABCD$ is a parallelogram

Statement Reason
1.
2. Vertical Angles Theorem
3. $\triangle AED \cong \triangle CEB\!\\\triangle AEB \cong \triangle CED$
4.
5. $ABCD$ is a parallelogram
1. Given: $\angle ADB \cong \angle CBD, \overline{AD} \cong \overline{BC}$ Prove: $ABCD$ is a parallelogram

Statement Reason
1.
2. $\overline{AD} \| \overline{BC}$
3. $ABCD$ is a parallelogram

1.

(a) $\text{Slope}\ AB = \text{Slope} \ CD = -\frac{1}{2}\!\\{\;}\quad \ \text{Slope}\ AD = \text{Slope}\ BC = \frac{2}{3}\\$

$ABCD$ is a parallelogram because the opposite sides are parallel.

(b) $\text{Midpoint of}\ BD = (0, -2)\!\\{\;}\quad \ \text{Midpoint of}\ AC = (0, -2)$

Yes, the midpoints of the diagonals are the same, so they bisect each other.

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Feb 22, 2012

Dec 11, 2014