# Lesson 6

Side-Angle-Side Triangle Congruence

### Problem 1

Triangle \(DAC\) is isosceles with congruent sides \(AD\) and \(AC\). Which additional given information is sufficient for showing that triangle \(DBC\) is isosceles? Select **all** that apply.

Line \(AB\) is an angle bisector of \(DAC\).

Angle \(BAD\) is congruent to angle \(ABC\).

Angle \(BDC\) is congruent to angle \(BCD\).

Angle \(ABD\) is congruent to angle \(ABC\).

Triangle \(DAB\) is congruent to triangle \(CAB\).

### Solution

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### Problem 2

Tyler has written an incorrect proof to show that quadrilateral \(ABCD\) is a parallelogram. He knows segments \(AB\) and \(DC\) are congruent. He also knows angles \(ABC\) and \(ADC\) are congruent. Find the mistake in his proof.

Segment \(AC\) is congruent to itself, so triangle \(ABC\) is congruent to triangle \(ADC\) by Side-Angle-Side Triangle Congruence Theorem. Since the triangles are congruent, so are the corresponding parts, and so angle \(DAC\) is congruent to \(ACB\). In quadrilateral \(ABCD\), \(AB\) is congruent to \(CD\) and \(AD\) is parallel to \(CB\). Since \(AD\) is parallel to \(CB\), alternate interior angles \(DAC\) and \(BCA\) are congruent. Since alternate interior angles are congruent, \(AB\) must be parallel to \(CD\). Quadrilateral \(ABCD\) must be a parallelogram since both pairs of opposite sides are parallel.

### Solution

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### Problem 3

Triangles \(ACD\) and \(BCD\) are isosceles. Angle \(BAC\) has a measure of 18 degrees and angle \(BDC\) has a measure of 48 degrees. Find the measure of angle \(ABD\).

### Solution

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### Problem 4

Here are some statements about 2 zigzags. Put them in order to prove figure \(ABC\) is congruent to figure \(DEF\).

- 1: If necessary, reflect the image of figure \(ABC\) across \(DE\) to be sure the image of \(C\), which we will call \(C'\), is on the same side of \(DE\) as \(F\).
- 2: \(C'\) must be on ray \(EF\) since both \(C'\) and \(F\) are on the same side of \(DE\) and make the same angle with it at \(E\).
- 3: Segments \(AB\) and \(DE\) are the same length so they are congruent. Therefore, there is a rigid motion that takes \(AB\) to \(DE\). Apply that rigid motion to figure \(ABC\).
- 4: Since points \(C'\) and \(F\) are the same distance along the same ray from \(E\) they have to be in the same place.
- 5: Therefore, figure \(ABC\) is congruent to figure \(DEF\).

### Solution

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(From Unit 2, Lesson 5.)### Problem 5

Match each statement using only the information shown in the pairs of congruent triangles.

### Solution

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(From Unit 2, Lesson 4.)### Problem 6

Triangle \(ABC\) is congruent to triangle \(EDF\). So, Priya knows that there is a sequence of rigid motions that takes \(ABC\) to \(EDF\).

Select **all** true statements after the transformations:

Segment \(AB\) coincides with segment \(EF\).

Segment \(BC\) coincides with segment \(DF\).

Segment \(AC\) coincides with segment \(ED\).

Angle \(A\) coincides with angle \(E\).

Angle \(C\) coincides with angle \(F\).

### Solution

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(From Unit 2, Lesson 3.)