E is a point inside square ABCD such that $\angle{ECD} = \angle{EDC} = 15.$ Find $\angle{AEB}.$
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1Does this answer your question? $ABCD$ is a square and $AEB$ is an equilateral triangle. Find $\angle DEC$. – nonuser Jun 18 '20 at 15:13
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Construct point $F$ in the interior of the square such that $\angle{FDA}=\angle{FAD}=15^{\circ}$.
$\textbf{Claim}$ $F$ is the circumcenter of $\triangle DEA$
To see this, observe that $FD=DE$ and that $\angle{FDE}=60^{\circ}$ by construction. This implies that $\triangle FDE$ is equilateral, from which the claim immediately follows.
A straightforward angle chase tells us that $\angle{FAE}=15^{\circ}$ and hence $\angle{EAB}=60^{\circ}$. We also have $EA=EB$ due to symmetry, and hence it follows that $\triangle EAB$ is equilateral. It immediately follows that $\angle EAB=60^{\circ}$.
Alan
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Let the side of the square be $a$. Now, applying Sine Rule in $\triangle EFD$, we get, $$DE=\frac{a}{2\sin 75°}.$$
Now, apply Sine Rule in $\triangle DEA$ $$\frac{\sin (15°+\theta)}{\cos \theta}=\frac{\frac{\sqrt 3+1}{2\sqrt 2}}{\frac{1}{2}}.$$
Clearly, $\theta=60°$ satisfies the equation. Now, you can find $\angle AEB$.
SarGe
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