I have got a problem which I have to solve for my practive for an exam. Hope you can help me.
An isosceles trapezoid $ABCD$ with the parallel sides $\overline{AB}$ and $\overline{CD}$ is given. The incircle of $\triangle BCD$ touches $\overline{CD}$ in the point $P$. The line perpendicular to $\overline{CD}$ at $P$ meets the bisector of $\angle DAC$ at $F$. The circumcircle of $ \triangle ACF$ cuts $\overleftrightarrow{CD}$ at $C$ and $G$.
Show that $\triangle AFG$ is isosceles.
Thank you.
Let $I$ be the incenter of $\triangle ACD$, let $E$ be the incenter of $\triangle BCD$, and let $Q$ be the foot of the altitude from $I$ to $CD$. Since $CP=QD$ by symmetry, it follows that $F$ is in fact the $A$-excenter of $\triangle ACD$. Then it follows that: $$\begin{align*}\angle CFE&=\frac{\pi}{2}-\angle FCD\\&=\frac{\angle ACD}{2}\\&=\frac{\angle ACD}{2}\\&=\frac{\angle BDC}{2} \\&=\angle EDC \end{align*}$$ Therefore $CEDF$ is cyclic. Finally, a second angle chase gives: $$\begin{align*}\angle FAG&=\angle FCG\\&=\angle FCD\\&=\angle FED\\&=\frac{\pi}{2}-\angle EDC\\&=\frac{\pi}{2}+\angle EDC-\angle BDC\\&=\pi-\angle FED-\angle BDC\\&=\pi-\angle DCF-\angle ACD\\&=\pi-\angle ACF\\&=\angle AGF\end{align*}$$ And so $\triangle AFG$ is isosceles with $\overline{AF}=\overline{FG}$ as required.