$\angle CDF=90^{\circ}-\angle FBD=\angle ABF=\angle AEF=180^{\circ}-\angle FEC$, hence $CEFD$ is cyclic. Thus $\angle AEH=180^{\circ}-\angle DEC=180^{\circ}-\angle DFC=\angle GFA$, so arc $AG$ and arc $AH$ have the same length, hence $AG=AH$.

Similar answer, well same but I liked practicing it We have, cyclic ABEF, then the angle AEF=angle ABF. We have AB perpendicular to CD and BF perpendicular to AD, the angle ABF=to the angle FDB is also fulfilled there they give us 3 equal angles therefore cyclic CDFE and since EGHF is cyclic, we have another 3 results of equal angles therefore CD||GH. And since AB passes through the center of the circumference it must be mediating GH, therefore AH = AG at the end of the exercise.