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# Quadrilaterals that are Parallelograms

## Ways to show if a quadrilateral has two pairs of parallel sides.

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Quadrilaterals that are Parallelograms

What if four friends, Geo, Trig, Algie, and Calc were marking out a baseball diamond? Geo is standing at home plate. Trig is 90 feet away at 3rd\begin{align*}3^{rd}\end{align*} base, Algie is 127.3 feet away at 2nd\begin{align*}2^{nd}\end{align*} base, and Calc is 90 feet away at 1st\begin{align*}1^{st}\end{align*} base. 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 and determine if the baseball diamond is a parallelogram.

### Quadrilaterals that are Parallelograms

Recall that a parallelogram is a quadrilateral with two pairs of parallel sides. Even if a quadrilateral is not marked with having two pairs of sides, it still might be a parallelogram. The following is a list of theorems that will help you decide if a quadrilateral is a parallelogram or not.

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

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

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

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

Each of these theorems can be a way to show that a quadrilateral is a parallelogram.

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. ABDCDB\begin{align*}\triangle ABD \cong \triangle CDB\end{align*} SSS
4. ABDBDC, ADBDBC\begin{align*}\angle ABD \cong \angle BDC, \ \angle ADB \cong \angle DBC\end{align*} CPCTC
5. AB¯¯¯¯¯¯¯¯ || DC¯¯¯¯¯¯¯¯, AD¯¯¯¯¯¯¯¯ || BC¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \ || \ \overline{DC},\ \overline{AD} \ || \ \overline{BC}\end{align*} Alternate Interior Angles Converse
6. ABCD\begin{align*}ABCD\end{align*} is a parallelogram Definition of a parallelogram

To show that a quadrilateral is a parallelogram in the xy\begin{align*}x-y\end{align*} plane, you will need to use a combination of the slope formulas, the distance formula and the midpoint formula. For example, to use the Definition of a Parallelogram, you would need to find the slope of all four sides to see if the opposite sides are parallel. To use the Opposite Sides Converse, you would have to find the length (using the distance formula) of each side to see if the opposite sides are congruent. To use the Parallelogram Diagonals Converse, you would need to use the midpoint formula for each diagonal to see if the midpoint is the same for both. Finally, you can use the last Theorem in this Concept (that if one pair of opposite sides is both congruent and parallel then the quadrilateral is a parallelogram) in the coordinate plane. To use this theorem, you would need to show that one pair of opposite sides has the same slope (slope formula) and the same length (distance formula).

#### Writing a Two-Column Proof

Write a two-column proof.

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

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

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

#### Recognizing Parallelograms

1. Is quadrilateral EFGH\begin{align*}EFGH\end{align*} a parallelogram? How do you know?

2. For part a, the opposite angles are equal, so by the Opposite Angles Theorem Converse, EFGH\begin{align*}EFGH\end{align*} is a parallelogram. In part b, the diagonals do not bisect each other, so EFGH\begin{align*}EFGH\end{align*} is not a parallelogram.

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

First, find the length of AB\begin{align*}AB\end{align*} and CD\begin{align*}CD\end{align*}.

AB=(13)2+(53)2=(4)2+22=16+4=20CD=(26)2+(2+4)2  =(4)2+22  =16+4  =20\begin{align*}AB& =\sqrt{(-1-3)^2+(5-3)^2} && CD=\sqrt{(2-6)^2+(-2+4)^2}\\ & = \sqrt{(-4)^2+2^2} && \quad \ \ =\sqrt{(-4)^2+2^2}\\ & = \sqrt{16+4} && \quad \ \ =\sqrt{16+4}\\ & = \sqrt{20} && \quad \ \ =\sqrt{20}\end{align*}

\begin{align*}AB = CD\end{align*}, so if the two lines have the same slope, \begin{align*}ABCD\end{align*} is a parallelogram.

Slope \begin{align*}AB = \frac{5-3}{-1-3}=\frac{2}{-4}=-\frac{1}{2}\end{align*} Slope \begin{align*}CD = \frac{-2+4}{2-6}=\frac{2}{-4}=-\frac{1}{2}\end{align*}

Therefore, \begin{align*}ABCD\end{align*} is a parallelogram.

#### Baseball Problem Revisited

First, we can 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*}

This means that the diagonals are equal. If the diagonals are equal, the other two sides of the diamond are also 90 feet. Therefore, the baseball diamond is a parallelogram.

### Examples

#### Example 1

What value of \begin{align*}x\end{align*} would make \begin{align*}ABCD\end{align*} a parallelogram?

\begin{align*}\overline{AB} \ || \ \overline{DC}\end{align*} from the markings. Therefore, \begin{align*}ABCD\end{align*} would be a parallelogram if \begin{align*}AB =DC\end{align*} as well.

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

In order for \begin{align*}ABCD\end{align*} to be a parallelogram, \begin{align*}x\end{align*} must equal 7.

#### Example 2

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

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

Midpoint of \begin{align*}RT = \left ( \frac{-4+3}{2}, \ \frac{3-4}{2} \right )=(-0.5,-0.5)\end{align*}

Midpoint of \begin{align*}SU = \left ( \frac{4-5}{2}, \ \frac{5-5}{2} \right ) =(-0.5,0)\end{align*}

Because the midpoint is not the same, \begin{align*}RSTU\end{align*} is not a parallelogram.

#### Example 3

If a quadrilateral has one pair of parallel sides is it a parallelogram?

Although it has one pair of parallel sides, this quadrilateral is not a parallelogram because its opposite sides are not necessarily congruent.

### Review

For questions 1-11, determine if the quadrilaterals are parallelograms. If they are, write a reason.

For questions 12-14, determine the value of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} that would make the quadrilateral a parallelogram.

For questions 15-17, 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*}

Write a two-column proof.

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
2. \begin{align*}{\;}\end{align*} Given: \begin{align*}\angle ADB \cong CBD, \ \overline{AD} \cong \overline{BC}\end{align*} Prove: \begin{align*}ABCD\end{align*} is a parallelogram

Suppose that \begin{align*}A(-2, 3), \ B(3, 3)\end{align*} and \begin{align*}C(1, -3)\end{align*} are three of four vertices of a parallelogram.

1. Depending on where you choose to put point \begin{align*}D\end{align*}, the name of the parallelogram you draw will change. Sketch a picture to show all possible parallelograms. How many can you draw?
2. If you know the parallelogram is named \begin{align*}ABDC\end{align*}, what is the slope of side parallel to \begin{align*}\overline{AC}\end{align*}?
3. Again, assuming the parallelogram is named \begin{align*}ABDC\end{align*}, what is the length of \begin{align*}\overline{BD}\end{align*}?

The points \begin{align*}Q(-1, 1), \ U(7, 1), \ A(1, 7)\end{align*} and \begin{align*}D(-1, 5)\end{align*} are the vertices of quadrilateral \begin{align*}QUAD\end{align*}. Plot the points on graph paper to complete problems 23-26.

1. Find the midpoints of sides \begin{align*}\overline{QU}, \ \overline{UA}, \ \overline{AD}\end{align*} and \begin{align*}\overline{DQ}\end{align*}. Label them \begin{align*}W, \ X, \ Y\end{align*} and \begin{align*}Z\end{align*} respectively.
2. Connect the midpoints to form quadrilateral \begin{align*}WXYZ\end{align*}. What does this quadrilateral appear to be?
3. Use slopes to verify your answer to problem 24.
4. Use midpoints to verify your answer to problem 24.

To view the Review answers, open this PDF file and look for section 6.4.

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