Question

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

Explanation:

In a proof by contradiction, we assume the opposite of what we want to prove (here, assume \( \angle B \) and \( \angle C \) are not congruent) and show it leads to a contradiction. The given is \( \overline{AB} \cong \overline{AC} \) (so \( AB = AC \)). Using the triangle parts relationship (larger angle opposite longer side), assuming \( \angle B \) and \( \angle C \) are not congruent leads to \( AC
eq AB \), contradicting \( AB = AC \). Thus, the assumption is false, so \( \angle B \cong \angle C \). The other options are incorrect: \( AB
eq AC \) contradicts the given; \( ABC \) is a triangle (given); \( \angle A \cong \angle B \cong \angle C \) isn't proven (only \( \angle B \cong \angle C \) follows from \( AB = AC \), making it isosceles, not necessarily equilateral).

Answer:

B. \( \angle B \cong \angle C \)