Problem: Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$.
The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$.
Let $F$ be a point on the (internal) angle bisector of $\angle DAC$
such that $EF \perp CD$.
Let the circumscribed circle of triangle $ACF$
meet line $CD$ at $C$ and $G$.
Prove that the triangle $AFG$ is isosceles.

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Solution: Note $E$ is contact point of $A$-excircle of $\triangle ACD$,
so $F$ is $A$-excenter.
Hence $CF$ is external angle bisector of $\angle ACG$
which implies $FA = FG$
(since $F$ is the arc midpoint on the circumcircle of $AFG$).