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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.

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    $\begingroup$ The bottom of your screenshot got cut off. Is there more to it? $\endgroup$
    – Calvin Lin
    Commented Feb 20, 2023 at 21:30

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Hints towards a solution.

  • 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.
  • Reverse this sequence of steps to get a direct solution.
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