# Find the angle $\angle CAE$

Let $$\triangle ABC$$ isosceles with $$AB=AC$$ such that there exists $$D$$ in its inside with $$AC=CD$$ and $$2\angle BDC + \angle BAC=360^{\circ}$$. Let $$E$$ the symmetry of $$D$$ relative to $$BC$$.

Find $$\angle CAE$$.

I think that this angle should be $$60^{\circ}$$. I tried to compute all the angles and find a congruence of triangles. I put below my approach.

• The bottom of your screenshot got cut off. Is there more to it? Commented Feb 20, 2023 at 21:30

• Work backwards assuming the answer is indeed $$60 ^ \circ$$. (Yes, this is a huge assumption, but let's say you are correct / I trust that you have looked at enough examples to guess the answer.)
• Then $$ACE$$ is an equilateral triangle. We already have $$CE = CD = CA$$ like you pointed out, so the unknown is $$AE$$.
• If $$AB = AE = AC$$, then $$A$$ is the circumcenter of triangle $$BEC$$.
• The angle condition supports that hypothesis, but we'd need a bit more than just that angle chasing to prove that $$A$$ is indeed the circumcenter. Fill in this gap.