Post Number:#2 by Thanic Nur Samin » Wed Jan 11, 2017 6:13 pm
It seems unbelievable that I missed this on the contest.
Using alternate segment theorem, it follows that $\triangle AEC ~ \triangle AFD$. From this, we see that $\angle AEF=\angle AFE$. Now, we get that $\angle APC=\angle AFD=\angle AEC$. However, $PE\perp AC$. So, $P$ is the reflection of $E$ over $AC$. Similarly, $Q$ is the reflection of $F$ over $AD$.
Now, the obvious part. $\angle PAQ=\angle PAC+\angle CAD+\angle DAQ=\angle ADC+\angle CAD+\angle ACD=180^{\circ}$. So, $P,A$ and $Q$ are collinear.
Also, I noticed that $AB$ is the symmedian of $\triangle ACD$. Is there any solution that lets us use that information?
#make_BdMO_forum_great_again