Romanian Mathematical Olympiad - Final Round

A straightforward angle chasing gives $\angle ABC = 90^\circ - 54^\circ = 36^\circ$, then $\angle ABD = \angle CBD = \frac{1}{2} \angle ABC = 18^\circ$. So $\angle BDA = 90^\circ - \angle ABD = 72^\circ$, $\angle BDC = 180^\circ - \angle BDA = 108^\circ$.

The isosceles triangle $CDE$ gives $\angle DCE = \angle DEC = \frac{1}{2}(180^\circ - \angle CDE) = 36^\circ$, whence $\angle BCE = \angle BCD - \angle ECD = 18^\circ = \angle CBE$, hence the triangle $BCE$ is isosceles.

Denote $F$ the reflection of $D$ into $A$. Then the points $D$, $A$, $F$ are collinear and $\angle ABF = \angle ABD = 18^\circ$ (because $\triangle BAF \equiv \triangle BAD$ – case S.A.S.), so $\angle FBC = 54^\circ = \angle FCB$, hence the triangle $BFC$ is isosceles, with $FB = FC$.

This gives $\triangle FEB \equiv \triangle FEC$ (S.S.S.), leading to $\angle EFB = \angle EFC = \frac{1}{2} \angle CFB = 36^\circ$. So triangle $EFC$ is isosceles, whence $EF = EC = EB$ (*). Also, $\angle FED = 180^\circ - \angle FDE - \angle FED = 180^\circ - 72^\circ - 36^\circ = 72^\circ = \angle FDE$, showing that $FE = FD = 2 \cdot AD$. Now (*) yields $EB = 2 \cdot AD$.