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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.

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  • $\begingroup$ I am really starting to like these geometry problems :) $\endgroup$ – Mann May 9 '15 at 11:44
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    $\begingroup$ So $G$ is $AF\cap CD$? In such a way, how can $AFG$ be possibly isosceles? $\endgroup$ – Jack D'Aurizio May 9 '15 at 12:28
  • $\begingroup$ No. The perimeter of ACF cuts the straight CD and the point which is different of C is G. $\endgroup$ – grok May 9 '15 at 12:31
  • $\begingroup$ @grok: Perhaps you mean the circumcircle of $\triangle ACF$, rather than the perimeter. (My GeoGebra sketch seems to show that point $G$ defined in this way makes an isosceles $\triangle AFG$.) $\endgroup$ – Blue May 9 '15 at 12:35
  • $\begingroup$ @Blue yes I mean the circumcircle but I thought that the perimeter is a synonym. I know that GeoGebra shows it but I have to prove it ;). And I need help because I don't know how I have to solve it. $\endgroup$ – grok May 9 '15 at 12:39
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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.

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  • $\begingroup$ thank you very much for your quick solution. But which point in your answer is the point E? ;) $\endgroup$ – grok May 10 '15 at 15:58
  • $\begingroup$ Ah, sorry about that! E is the incenter of $\triangle BCD$. I will edit. $\endgroup$ – Apple May 10 '15 at 19:00

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