# Lesson 15

Congruence for Quadrilaterals

### Problem 1

Select **all** quadrilaterals that have 180 degree rotational symmetry.

trapezoid

isosceles trapezoid

parallelogram

rhombus

rectangle

square

### Solution

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(From Unit 2, Lesson 14.)### 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 \(\underline{\hspace{.5in}1\hspace{.5in}}\) is the same as the distance from \(E\) to \(\underline{\hspace{.5in}2\hspace{.5in}}\). Since \(E\) is the same distance from \(\underline{\hspace{.5in}3\hspace{.5in}}\) as it is from \(\underline{\hspace{.5in}4\hspace{.5in}}\), it must lie on the perpendicular bisector of segment \(\underline{\hspace{.5in}5\hspace{.5in}}\). By the same reasoning, \(G\) must lie on the perpendicular bisector of \(\underline{\hspace{.5in}6\hspace{.5in}}\). Therefore, line \(\underline{\hspace{.5in}7\hspace{.5in}}\) is the perpendicular bisector of segment \(FH\). So reflecting rhombus \(EFGH\) across line \(\underline{\hspace{.5in}8\hspace{.5in}}\) will take \(E\) to \(\underline{\hspace{.5in}9\hspace{.5in}}\) and \(G\) to \(\underline{\hspace{.5in}10\hspace{.5in}}\) (because \(E\) and \(G\) are on the line of reflection) and \(F\) to \(\underline{\hspace{.5in}11\hspace{.5in}}\)and \(H\) to \(\underline{\hspace{.5in}12\hspace{.5in}}\) (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 \(EG\) must be a line of symmetry for rhombus \(EFGH\).

### Solution

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(From Unit 2, Lesson 14.)### 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 \(\underline{\hspace{.5in}1\hspace{.5in}}\), alternate interior angles \(\underline{\hspace{.5in}2\hspace{.5in}}\) and \(\underline{\hspace{.5in}3\hspace{.5in}}\) are congruent. \(AC\) is congruent to \(\underline{\hspace{.5in}4\hspace{.5in}}\) since segments are congruent to themselves. Along with the given information that \(AD\) is congruent to \(BC\), triangle \(ADC\) is congruent to \(\underline{\hspace{.5in}5\hspace{.5in}}\)by the \(\underline{\hspace{.5in}6\hspace{.5in}}\) Triangle Congruence. Since the triangles are congruent, all pairs of corresponding angles are congruent, so angle \(DCA\) is congruent to \(\underline{\hspace{.5in}7\hspace{.5in}}\). Since those alternate interior angles are congruent, \(AB\) must be parallel to \(\underline{\hspace{.5in}8\hspace{.5in}}\). Since we define a parallelogram as a quadrilateral with both pairs of opposite sides parallel, \(ABCD\) is a parallelogram.

### Solution

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(From Unit 2, Lesson 13.)### Problem 4

Select the statement that **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.

### Solution

### Problem 5

\(EFGH\) is a parallelogram and angle \(HEF\) is a right angle. Select **all **statements that **must **be true.

\(EFGH \) is a rectangle.

Triangle \(HEF\) is congruent to triangle \(GFH\).

Triangle \( HEF\) is congruent to triangle \(FGH\).

\(ED \) is congruent to \(HD\), \(DG\), and \(DF\).

Triangle \(EDH\) is congruent to triangle \(HDG\).

### Solution

### Problem 6

Figure \(ABCD\) is a parallelogram. Is triangle \(ADB\) congruent to triangle \(CBD\)? Show or explain your reasoning.

### Solution

### Problem 7

Figure \(KLMN\) is a parallelogram. Prove that triangle \(KNL\) is congruent to triangle \(MLN\).