Question

ven: $\triangle abc$ is isosceles; $\overline{ab} \cong \overline{ac}$
ove: $\angle b \cong \angle c$
we are given that $\triangle abc$ is isosceles with $\overline{ab} \cong \overline{ac}$. using the definition of congruent line segments, we know that $ab = ac$.
lets assume that angles b and c are not congruent. then one angle measure must be greater than the other. if $m\angle b$ is greater than $m\angle c$, then $ac$ is greater than $ab$ by the triangle parts relationship theorem.
however, this contradicts the given information that dropdown options: side ab is congruent to side bc, side ac is congruent to side bc, side ab is congruent to side ac. therefore, which is what we rest of text cut off c, we would reach refore, the angles

Explanation:

The problem is about proving that in an isosceles triangle \( \triangle ABC \) with \( \overline{AB} \cong \overline{AC} \), \( \angle B \cong \angle C \). The proof uses a proof by contradiction. We start with the given that \( AB = AC \) (from congruent segments). Then we assume \( \angle B \) and \( \angle C \) are not congruent, which leads to a contradiction with the given \( AB \cong AC \) (since the triangle parts relationship theorem would imply a different side length relationship). The contradiction is because we assumed the opposite of what we know ( \( AB \cong AC \) ), so the correct option to fill in the contradiction part is the one that restates the given congruence of \( AB \) and \( AC \).

Answer:

side AB is congruent to side AC