$GC$ intersects with the circumscribed circle at $K$ , so $\angle BAK=\angle DAK$ So it is easy to prove $\triangle AEB \sim\triangle AGK$ and thus by angle-chasing $F,E,C,G$ is concyclic. Note that $\angle AFG=180-\angle AEG=180-\angle ACG=\angle ACK=\angle ADK$ Thus $\triangle ADK\sim\triangle AFG$ The two pairs of similar triangles are enough to prove the result.

At first we claim that $A,E,C,G$ is cyclic.Indeed, $\angle ECG=\angle 90^{\circ}+\frac{\angle C}{2}$ and $\angle EAG=\frac{\angle BAD}{2}=90^{\circ}-\frac{C}{2}$. Then,we claim that $\angle AGF= \angle ABD$.Indeed, \begin{align*} \angle AGF&=\angle GFE-\angle FAG\\ \\ &=\angle GEF-(90^{\circ}\frac{\angle C}{2})\\ &=\angle GCA-\angle GCD\\ &=\angle DCA \end{align*}Now observe that, \begin{align*} \frac{AB}{AE}&=\frac{\sin \angle AEC}{\sin \angle ABE}\\ &=\frac{\sin \angle AEC}{\sin \angle ADC}\\ &=\frac{\frac{AC}{AE} \sin \angle ACE}{\frac{AC}{AD}\sin \angle DCA}\\ &=\frac{AD}{AE}\frac{\sin \angle AGE}{\sin \angle DCA}\\ &=\frac{AD}{AE}\frac{\sin \angle AGE}{\sin \angle AGF}\\ &=\frac{AD}{AF} \frac{AF}{AE} \frac{\sin \angle AGE}{\sin \angle AGF}\\ &=\frac{AD}{AF} \frac{\frac{GF}{\sin \angle GAF} \sin\angle AGF}{\frac{GE}{\sin \angle GAE}\sin \angle GAE} \frac{\sin \angle AGE}{\sin \angle AGF}\\ &=\frac{AD}{AF} \end{align*}

As a P2 on the second day, it should have been harder(like the one in 2018 or 2017). My solution during the contest:

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