E is a point inside square $ABCD$ such that $ECD = EDC = 15$... E is a point inside square ABCD such that $\angle{ECD} = \angle{EDC} = 15.$ Find $\angle{AEB}.$
I drew a picture for this but I don't know how to continue. Any help is appreciated.


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 $\angle AEB$.
