Saudi Arabia Mathematical Competitions 2012

Let $\widehat{ABC}=2\beta$, $\widehat{BAC}=2\alpha$, $\widehat{ACB}=2\gamma$. Thus, the law of sines on triangle $AEF$ gives us

$$\frac{AE}{AF}=\frac{\sin (\alpha +2\beta )}{\sin (\alpha -\beta +\gamma )}.$$

On the other hand, using the law of sines on $ABD$, $ADF$ and the fact that $\widehat{BAD}=180^{\circ }-\widehat{DAF}$, we get that

$$\frac{AB}{AF}=\frac{BD\sin (\widehat{BDA})}{DF\sin (\widehat{FDA})}=\frac{BD\sin (\alpha +2\beta )}{DF\sin (\alpha -\beta +\gamma )}.$$

Combining these two identities, we obtain that $\frac{AE}{AB}=\frac{DF}{DB}$. Then since we also have $\widehat{FDB}=\widehat{FAE}=\widehat{EAB}$, triangle $EAB$ is similar to triangle $FDB$. Thus we find that $\widehat{FBE}=2\beta$. Using this and the fact that $A,D,F,E,G$ are on the same circle, we are able to compute that $\widehat{GEA}=\alpha -\beta +\gamma$. But this was also computed to be equal to $\widehat{AEF}$. Then $\widehat{GFA}=\widehat{GEA}=\widehat{AEF}=\widehat{AGF}$, so $AF=AG$ as desired.