Question

consider the diagram and proof by contradiction. given: △abc with (overline{ab}congoverline{ac}). since it is given that (overline{ab}congoverline{ac}), it must also be true that ab = ac. assume ∠b and ∠c are not congruent. then the measure of one angle is greater than the other. if (mangle b>mangle c), then ac > ab because of the triangle parts relationship theorem. for the same reason, if (mangle b < mangle c), then ac < ab. this is a contradiction to what is given. therefore, it can be concluded that
ab≠ac
∠b≅∠c
abc is not a triangle
∠a≅∠b≅∠c

Explanation:

Step1: Analyze the proof - by - contradiction

We start with the given $\overline{AB}\cong\overline{AC}$ (i.e., $AB = AC$). Then we assume $\angle B$ and $\angle C$ are not congruent. By the triangle - parts relationship theorem, if $m\angle B>m\angle C$, then $AC > AB$, and if $m\angle B

Step2: Draw the conclusion

Since the assumption that $\angle B$ and $\angle C$ are not congruent leads to a contradiction, the opposite must be true. That is, $\angle B\cong\angle C$.

Answer:

B. $\angle B\cong\angle C$