For the triangles to be congruent, we'd need the corresponding segments to be congruent, so DE would need to be congruent to AC, not BC. Two of its angles are congruent to ∆ ABC (∠ D ≅ ∠ A and ∠ E ≅ ∠ C), and the segment in the middle is also congruent to one of the segments of ∆ ABC ( DE ≅ BC). So let's see if their sides and angles match up. If their sides and angles match up, they're congruent. It says: If we're given ∆ ABC, AB ≅ BC if and only if ∠ BAC ≅ ∠ BCA. Since we've now proved this idea inside and out, we can finally give it a name: the Isosceles Triangle Theorem. The angles opposite their congruent sides are also congruent. What we've done is prove something pretty darn important about isosceles triangles. We don't need to, but we'll give you the formal proof just in case. Since both the triangles are congruent, all their sides are congruent. So what do we have? Two angles and a side in between them for both triangles-each one congruent to the other triangle's corresponding part. The side shared by both triangles is definitely congruent to itself. The given angles, ∠ BAC and ∠ ACB, are congruent. The angle bisector will make two congruent angles, one in each smaller triangle. That'll split ∆ ABC into two triangles with BD as a shared side. What we can do is bisect the unknown angle. What do these proofs want from us and why can't they just leave us alone? Buck up, buddy. Remember how we proved that isosceles triangles have two congruent angles because they have two congruent sides? This proof is asking us to do the exact opposite. Prove: ∆ ABC is isosceles because AB ≅ BC.
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