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6.3: Proving Quadrilaterals are Parallelograms

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

  • Prove a quadrilateral is a parallelogram.
  • Show a quadrilateral is a parallelogram in the xy 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 AB¯¯¯¯¯¯¯¯,BC¯¯¯¯¯¯¯¯,CD¯¯¯¯¯¯¯¯, and 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. 3rd base is 90 feet away, 2nd base is 127.3 feet away, and 1st base is also 90 feet away. The angle at home plate is 90, from 1st to 3rd is 90. 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¯¯¯¯¯¯¯¯

Prove: ABCD is a parallelogram

Statement Reason
1. AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯,AD¯¯¯¯¯¯¯¯BC¯¯¯¯¯¯¯¯ Given
2. DB¯¯¯¯¯¯¯¯DB¯¯¯¯¯¯¯¯ Reflexive PoC
3. ABDCDB SSS
4. ABDBDC,ADBDBC CPCTC
5. AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯,AD¯¯¯¯¯¯¯¯BC¯¯¯¯¯¯¯¯ Alternate Interior Angles Converse
6. ABCD is a parallelogram Definition of a parallelogram

Example 1: Write a two-column proof.

Given: AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯, and AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯

Prove: ABCD is a parallelogram

Solution:

Statement Reason
1. AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯, and AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯ Given
2. ABDBDC Alternate Interior Angles
3. DB¯¯¯¯¯¯¯¯DB¯¯¯¯¯¯¯¯ Reflexive PoC
4. ABDCDB SAS
5. AD¯¯¯¯¯¯¯¯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: AB¯¯¯¯¯¯¯¯DC¯¯¯¯¯¯¯¯. By Theorem 6-10, ABCD would be a parallelogram if AB=DC.

5x83xx=2x+13=21=7

Showing a Quadrilateral is a Parallelogram in the xy Plane

To show that a quadrilateral is a parallelogram in the xy plane, you might need:

  • The Slope Formula, y2y1x2x1.
  • The Distance Formula, (x2x1)2+(y2y1)2.
  • The Midpoint Formula, (x1+x22,y1+y22).

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=(13)2+(53)2=(4)2+22=16+4=20CD=(26)2+(2+4)2=(4)2+22=16+4=20

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*}
3. Definition of a quadrilateral
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

Review Queue Answers

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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