Question

given: $overline{ar}$ bisects $angle bac$; $ab = ac$ which congruence theorem can be used to prove $\triangle abr cong \triangle acr$? $\bigcirc$ aas $\bigcirc$ sss $\bigcirc$ asa $\bigcirc$ sas

Explanation:

1. Recall the given information: \( \overline{AR} \) bisects \( \angle BAC \), so \( \angle BAR = \angle CAR \) (definition of angle bisector). Also, \( AB = AC \) (given), and \( \overline{AR} \) is a common side to both \( \triangle ABR \) and \( \triangle ACR \), so \( AR = AR \) (reflexive property of congruence).
2. Now, identify the parts of the triangles: In \( \triangle ABR \) and \( \triangle ACR \), we have:

• \( AB = AC \) (given, one side).
• \( \angle BAR = \angle CAR \) (from angle bisector, one included angle).
• \( AR = AR \) (common side, another side).

1. The SAS (Side - Angle - Side) congruence theorem states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent. Here, we have two sides (\( AB = AC \), \( AR = AR \)) and the included angle (\( \angle BAR=\angle CAR \)) congruent, so the SAS theorem applies.

• AAS (Angle - Angle - Side) requires two angles and a non - included side, which is not the case here.
• SSS (Side - Side - Side) requires three sides, but we only know two sides and an included angle.
• ASA (Angle - Side - Angle) requires two angles and the included side, but here we have two sides and the included angle, not two angles and the included side in the ASA sense.

Answer:

\(\boldsymbol{\text{SAS}}\) (the option corresponding to SAS, e.g., if the options are labeled as in the problem, the answer is the option with "SAS")