Please correct me if I'm wrong. It's obvious that BECF and CFHD are cyclic. From GD||AB we have $\angle GDH=90 - $\angle ABC= $\angle GHF so points C, H, D, G, F are concyclic and $\angle CGH=90

MathsLion wrote: Please correct me if I'm wrong. It's obvious that $BECF$ and $CFHD$ are cyclic. From $GD||AB$ we have $\angle GDH=90 - \angle ABC$= $\angle GHF$ so points $C, H, D, G, F$ are concyclic and $\angle CGH=90$

$$\angle BAD=\angle DEB=90^{\circ}-B=\angle ADG \text{ and } \angle HEG=180^{\circ}-(90^{\circ}-B)=90^{\circ}+B \implies DHEG \text{ is cyclic or } H \text{ passes through } (DEG)$$Now, Since, $$ AB||DG \text{ and Let } DG \cap CF=X \implies \angle FXG=90^{\circ} \implies \angle FGX=\angle C=\angle ECD \implies DEGC \text{ is cyclic or } C \text{ passes though } (DEG)$$Hence, $DHEGC$ is cyclic $\implies \angle HEC=\boxed{\angle CGH=90^{\circ}}$

Accually, the quadrilaterals ${HDFG}$ and ${HDCF}$ are cyclic, so the quadrilateral ${HDCG}$ is also cyclic. Finally, since ${\angle CDH=90^{\circ}}$ it will be ${\angle CGH=90^{\circ}}$