0
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ I am really starting to like these geometry problems :) $\endgroup$ – Mann May 9 '15 at 11:44
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.