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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 xy\begin{align*}x-y\end{align*} plane.

Review Queue

1. Plot the points A(2,2),B(4,2),C(2,4)\begin{align*}A(2, 2), B(4, -2), C(-2, -4)\end{align*}, and D(6,2)\begin{align*}D(-6, -2)\end{align*}.
1. Find the slopes of AB¯¯¯¯¯¯¯¯,BC¯¯¯¯¯¯¯¯,CD¯¯¯¯¯¯¯¯,\begin{align*}\overline{AB},\overline{BC}, \overline{CD},\end{align*} and AD¯¯¯¯¯¯¯¯\begin{align*}\overline{AD}\end{align*}. Is ABCD\begin{align*}ABCD\end{align*} 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. 3rd\begin{align*}3^{rd}\end{align*} base is 90 feet away, 2nd\begin{align*}2^{nd}\end{align*} base is 127.3 feet away, and 1st\begin{align*}1^{st}\end{align*} base is also 90 feet away. The angle at home plate is 90\begin{align*}90^\circ\end{align*}, from 1st\begin{align*}1^{st}\end{align*} to 3rd\begin{align*}3^{rd}\end{align*} is 90\begin{align*}90^\circ\end{align*}. 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: AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯,AD¯¯¯¯¯¯¯¯BC¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{DC}, \overline{AD} \cong \overline{BC}\end{align*}

Prove: ABCD\begin{align*}ABCD\end{align*} is a parallelogram

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

Example 1: Write a two-column proof.

Given: \begin{align*}\overline{AB} \| \overline{DC}\end{align*}, and \begin{align*}\overline{AB} \cong \overline{DC}\end{align*}

Prove: \begin{align*}ABCD\end{align*} is a parallelogram

Solution:

Statement Reason
1. \begin{align*}\overline{AB} \| \overline{DC}\end{align*}, and \begin{align*}\overline{AB} \cong \overline{DC}\end{align*} Given
2. \begin{align*}\angle ABD \cong \angle BDC\end{align*} Alternate Interior Angles
3. \begin{align*}\overline{DB} \cong \overline{DB}\end{align*} Reflexive PoC
4. \begin{align*}\triangle ABD \cong \triangle CDB\end{align*} SAS
5. \begin{align*}\overline{AD} \cong \overline{BC}\end{align*} CPCTC
6. \begin{align*}ABCD\end{align*} 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 \begin{align*}EFGH\end{align*} a parallelogram? How do you know?

Solution:

a) By the Opposite Angles Theorem Converse, \begin{align*}EFGH\end{align*} is a parallelogram.

b) \begin{align*}EFGH\end{align*} is not a parallelogram because the diagonals do not bisect each other.

Example 3: Algebra Connection What value of \begin{align*}x\end{align*} would make \begin{align*}ABCD\end{align*} a parallelogram?

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

\begin{align*}5x - 8 & = 2x + 13\\ 3x & = 21\\ x & = 7\end{align*}

Showing a Quadrilateral is a Parallelogram in the \begin{align*}x-y\end{align*} Plane

To show that a quadrilateral is a parallelogram in the \begin{align*}x-y\end{align*} plane, you might need:

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

Example 4: Is the quadrilateral \begin{align*}ABCD\end{align*} a parallelogram?

Solution: Let’s use Theorem 6-10 to see if \begin{align*}ABCD\end{align*} is a parallelogram. First, find the length of \begin{align*}AB\end{align*} and \begin{align*}CD\end{align*}.

\begin{align*}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}\end{align*}

Find the slopes.

\begin{align*}\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}\end{align*}

\begin{align*}AB = CD\end{align*} and the slopes are the same, \begin{align*}ABCD\end{align*} is a parallelogram.

Example 5: Is the quadrilateral \begin{align*}RSTU\end{align*} a parallelogram?

Solution: Let’s use the Parallelogram Diagonals Converse to see if \begin{align*}RSTU\end{align*} is a parallelogram. Find the midpoint of each diagonal.

\begin{align*}&\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)\end{align*}

\begin{align*}RSTU\end{align*} 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.

\begin{align*}90^2 + 90^2 & = d^2\\ 8100 + 8100 & = d^2\\ 16200 & = d^2\\ d & = 127.3\end{align*}

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 \begin{align*}x\end{align*} and \begin{align*}y\end{align*} that would make the quadrilateral a parallelogram.

For questions 19-22, determine if \begin{align*}ABCD\end{align*} is a parallelogram.

1. \begin{align*}A(8, -1), B(6, 5), C(-7, 2), D(-5, -4)\end{align*}
2. \begin{align*}A(-5, 8), B(-2, 9), C(3, 4), D(0, 3)\end{align*}
3. \begin{align*}A(-2, 6), B(4, -4), C(13, -7), D(4, -10)\end{align*}
4. \begin{align*}A(-9, -1), B(-7, 5), C(3, 8), D(1, 2)\end{align*}

Fill in the blanks in the proofs below.

1. Opposite Angles Theorem Converse

Given: \begin{align*}\angle A \cong \angle C, \angle D \cong \angle B\end{align*}

Prove: \begin{align*}ABCD\end{align*} is a parallelogram

Statement Reason
1.
2. \begin{align*}m \angle A = m \angle C, m \angle D = m \angle B\end{align*}
4. \begin{align*}m \angle A + m \angle A + m \angle B + m \angle B = 360^\circ\end{align*}
5. Combine Like Terms
6. Division PoE

7. \begin{align*}\angle A\end{align*} and \begin{align*}\angle B\end{align*} are supplementary

\begin{align*}\angle A\end{align*} and \begin{align*}\angle D\end{align*} are supplementary

8. Consecutive Interior Angles Converse
9. \begin{align*}ABCD\end{align*} is a parallelogram
1. Parallelogram Diagonals Theorem Converse

Given: \begin{align*}\overline{AE} \cong \overline{EC}, \overline{DE} \cong \overline{EB}\end{align*}

Prove: \begin{align*}ABCD\end{align*} is a parallelogram

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

Statement Reason
1.
2. \begin{align*}\overline{AD} \| \overline{BC}\end{align*}
3. \begin{align*}ABCD\end{align*} is a parallelogram

1.

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

\begin{align*}ABCD\end{align*} is a parallelogram because the opposite sides are parallel.

(b) \begin{align*}\text{Midpoint of}\ BD = (0, -2)\!\\ {\;}\quad \ \text{Midpoint of}\ AC = (0, -2)\end{align*}

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

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