Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$ and $AD$ intersect the circumcircle of $ABC$ for the second time at $E$. Let $P$ be the point symmetric to the point $E$ with respect to the point $D$ and $Q$ be the point of intersection of the lines $CP$ and $AB$. Prove that if $A,C,D,Q$ are concyclic, then the lines $BP$ and $AC$ are perpendicular.
Problem
Source: Turkey EGMO TST 2014 P1
Tags: geometry, circumcircle
lebathanh
07.08.2016 16:38
any solution
trungnghia215
07.08.2016 17:30
Since $P$ is midpoint of both $BC$ and $PE$ then $BECP$ is a parallelogram. Hence $\angle CBE = \angle BCP = \angle DCQ = \angle DAQ = \angle EAB$. Thus, $AE$ is internal bisector of $\angle A$ or $\triangle ABC$ is a isosceles triangle. Then the result follows as desired.
hakN
26.04.2021 23:25
If $A,C,D,Q$ are concyclic, then we have $\angle BAE = \angle QAD = \angle QCD = \angle CBE = \angle CAE$, implying that $AM$ is also angle bisector of $\angle BAC$ and so $AB=AC$. So $90 = \angle ADC = \angle AQC$ and so $P$ is the orthocenter of $\triangle ABC$ and the result follows.
primesarespecial
10.08.2021 20:52
Meh,wasn't this simple for EGMO, $BECP$ is a parallelogram,Now, $\angle CAE=\angle EBC=\angle QCD=\angle QAD$, $ABC$ is isosceles with $P$ are orthocentre. No motivation was required for this problem,I guess, but u could say that it is known that if $BP$ was perp then we would have $ABC$ isosceles,so...