CDEF is a square, CBF and BEA are equilateral triangles. Find angle $x$.
What I have found at the moment:
- $\angle BEF =15^{\circ}, \angle AED =135^{\circ} $
- I managed to find the length of $BE$ and use the Law of Cosines in $\triangle ADE$ to find $AD$ and to use the Law of Cosines one more time to get the $\cos \angle ADE$, but I didn't get good numbers.
Maybe there should be an analytical way to solve the problem.
Simple angle chasing leads to, $$\angle CBD=\angle CDB=\angle EFB=\angle EBF=15^{\circ}$$ Observe, $$\triangle BCD\cong \triangle BEF\implies BD=BF=AB\implies \angle BDA=\angle BAD=45^{\circ}$$ Since $\angle CDB=15^{\circ}$ and $\angle BDF=\angle BFD=75^{\circ},$ $$\angle BDA=75^{\circ}-x\implies \boxed{x=30^{\circ}}$$