# Congruence for Quadrilaterals

23 questions

# Congruence for Quadrilaterals

##### Problem 1

1) Select $\textbf{all}$ quadrilaterals that have 180 degree rotational symmetry. Write each corresponding letter in the answer box and separate letters with commas.

a) trapezoid $\quad\quad$ b) isosceles trapezoid $\quad\quad$ c) parallelogram $\quad\quad$ d) rhombus $\quad\quad$ e) rectangle $\quad\quad$ f) square

##### Problem 2

Lin wrote a proof to show that diagonal $EG$ is a line of symmetry for rhombus $EFGH$. Fill in the blanks to complete her proof:

Because $EFGH$ is a rhombus, the distance from $E$ to $\underline{\quad\quad 1 \quad\quad}$ is the same as the distance from $E$ to $\underline{\quad\quad 2 \quad\quad}$. Since $E$ is the same distance from $\underline{\quad\quad 3 \quad\quad}$ as it is from $\underline{\quad\quad 4 \quad\quad}$, it must lie on the perpendicular bisector of segment $\underline{\quad\quad 5 \quad\quad}$. By the same reasoning, $G$ must lie on the perpendicular bisector of $\underline{\quad\quad 6 \quad\quad}$. Therefore, line $\underline{\quad\quad 7 \quad\quad}$ is the perpendicular bisector of segment $FH$. So reflecting rhombus $EFGH$ across line $\underline{\quad\quad 8 \quad\quad}$ will take $E$ to $\underline{\quad\quad 9 \quad\quad}$ and $G$ to $\underline{\quad\quad 10 \quad\quad}$ (because $E$ and $G$ are on the line of reflection) and $F$ to $\underline{\quad\quad 11 \quad\quad}$ and $H$ to $\underline{\quad\quad 12 \quad\quad}$ (since $FH$ is perpendicular to the line of reflection, and $F$ and $H$ are the same distance from the line of reflection, on opposite sides). Since the image of rhombus $EFGH$ reflected across $EG$ is rhombus $EHGF$ (the same rhombus!), line $EF$ must be a line of symmetry for rhombus $EFGH$.

Fill in the blanks using items from the Bank of Terms below. Some items might be used more than once.

Bank of Terms: $E$, $F$, $G$, $H$, $EG$, $FH$

2) Blanks 1 and 2, separated by a comma

3) Blanks 3 and 4, separated by a comma

4) Blank 5

5) Blank 6

6) Blank 7

7) Blank 8

8) Blank 9

9) Blank 10

10) Blank 11

11) Blank 12

##### Problem 3

In quadrilateral $ABCD$, $AD$ is congruent to $BC$, and $AD$ is parallel to $BC$. Andre has written a proof to show that $ABCD$ is a parallelogram. Fill in the blanks to complete the proof.

Since $AD$ is parallel to $\underline{\quad\quad 1 \quad\quad}$, alternate interior angles $\underline{\quad\quad 2 \quad\quad}$ and $\underline{\quad\quad 3 \quad\quad}$ are congruent. $AC$ is congruent to $\underline{\quad\quad 4 \quad\quad}$ since segments are congruent to themselves. Along with the given information that $AD$ is congruent to $BC$, triangle $ADC$ is congruent to$\underline{\quad\quad 5 \quad\quad}$ by the $\underline{\quad\quad 6 \quad\quad}$ Triangle Congruence. Since the triangles are congruent, all pairs of corresponding angles are congruent, so angle $DCA$ is congruent to $\underline{\quad\quad 7 \quad\quad}$. Since those alternate interior angles are congruent, $AB$ must be parallel to $\underline{\quad\quad 8 \quad\quad}$. Since we define a parallelogram as a quadrilateral with both pairs of opposite sides parallel, $ABCD$ is a parallelogram.

Fill in the blanks using items from the Bank of Terms below. Some items might be used more than once.

Bank of Terms: $AC$, $BC$, $CD$, $BAC$, $BCA$, $CBA$, $DAC$, Side-Angle-Side

12) Blank 1

13) Blanks 2 and 3, separated by a comma

14) Blank 4

15) Blank 5

16) Blank 6

17) Blank 7

18) Blank 8

##### Problem 4

19) Select the statement that $\textbf{must}$ be true.

Parallelograms have at least one right angle.

If a quadrilateral has opposite sides that are both congruent and parallel, then it is a parallelogram.

Parallelograms have congruent diagonals.

The height of a parallelogram is greater than the lengths of the sides.

##### Problem 5

20) $EFGH$ is a parallelogram and angle $HEF$ is a right angle. Select $\textbf{all}$ statements that $\textbf{must}$ be true. Write each corresponding letter in the answer box and separate letters with commas.

a) $EFGH$ is a rectangle. $\quad\quad$ b) Triangle $HEF$ is congruent to triangle $GFH$. $\quad\quad$ c) Triangle $HEF$ is congruent to triangle $FGH$. $\quad\quad$ d) $ED$ is congruent to $HD$, $DG$, and $DF$. $\quad\quad$ e) Triangle $EDH$ is congruent to triangle $HDG$.

##### Problem 6

Figure $ABCD$ is a parallelogram.

Given: $\overline{AB} \cong \overline{CD}$, $\angle ADB \cong \angle CBD$

21) Is triangle $ADB$ congruent to triangle $CBD$?

22) Explain your reasoning.

##### Problem 7

23) Figure $KLMN$ is a parallelogram. Prove that triangle $KNL$ is congruent to triangle $MLN$.