Angle Triangulated

Construction: Rotate $\triangle ABE$ about point $A$ by an angle of $90^{\circ}$ , to form $\triangle ADE'$ $(B\to D,\; E\to E')$ In $\triangle AFE$ and $\triangle AFE'$ , $\begin{aligned} \angle EAF=90^{\circ}-\angle DAF-\angle BAE=45^{\circ}, \;\; &\angle FAE'=\angle DAF+\angle DAE'=45^{\circ}, \;\; AE=AE' \\ \end{aligned}$ $\therefore \triangle AFE\cong \triangle AFE' \implies \angle AEF=\angle AE'D=\angle AEB=72^{\circ}$ Since $\angle HGI=\angle ICH=90^{\circ}$ , quadrilateral $GHCI$ is cyclic , $\angle GCH=\angle GIH=\alpha$ Also, $G$ is the midpoint of the hypotenuse ( $EF$ ) of right-triangle $ECF$ , $GF=GE=GC \implies \angle GEC=\angle GCH=\alpha$ The points $B, E, C$ lie on a straight line, $\begin{aligned} \angle BEA+\angle AEF+\angle CEF&=180^{\circ} \\ 72^{\circ}+72^{\circ}+\alpha&=180^{\circ} \\ \implies \boxed{\alpha=36^{\circ}} \end{aligned}$