Team selection tests for IMO 2018

Let $D$ be the second intersection of $AP$ and $(O)$ and $H$ be the intersection of $OD$ and $AE$. Note that $D$ is the midpoint of the minor arc $BC$ of $(O)$, then $OD$ is the perpendicular bisector of $BC$. Since $AE\parallel BC$, we have $\angle DHE=90^{\circ }$.



On the other hand, $$\angle DOG=2\angle DAG=\angle PKG,$$ which implies that two isosceles triangles $KPG$ and $ODG$ are similar. We get $\angle ODG=\angle KPG$ therefore $PFGD$ is cyclic.

From this, $\angle AEG=\angle DPG=\angle DFG$. This means $HEGF$ is cyclic, we deduce $\angle EGF=180^{\circ }-\angle EHF=90^{\circ }$.