Hidden Configuration of Symmedians, Beautiful!... INAMO 2015 Shortist G6 wrote: Let $ABC$ be an acute angled triangle with circumcircle $O$. Line $AO$ intersects the circumcircle of triangle $ABC$ again at point $D$. Let $P$ be a point on the side $BC$. Line passing through $P$ perpendicular to $AP$ intersects lines $DB$ and $DC$ at $E$ and $F$ respectively . Line passing through $D$ perpendicular to $BC$ intersects $EF$ at point $Q$. Prove that $EQ = FQ$ if and only if $BP = CP$. Solution:- Claim 1:- $AEDF$ is a cyclic quadrialteral. $\angle ABC=\angle AEF=\angle ADF\implies AEDF$ is a cyclic quadrilateral. Claim 2:- Moreover $AEDF$ is a harmonic quadrilateral. Notice that due to the Similarity of the triangles $ABE$ and $ACF$ we get that $\frac{AB}{AC}=\frac{AE}{AF}=\frac{BE}{CF}$. Now by Menelaus Theorem we get that $$\frac{DF}{FC}\times\frac{CP}{PB}\times\frac{BE}{ED}=1\implies \frac{DF}{FC}=\frac{ED}{EB}\implies \frac{EB}{FC}=\frac{ED}{DF}=\frac{AB}{AC}=\frac{AE}{AF}\implies (A,D;E,F)=-1\implies AEDF \text{is a harmonic quadrilateral}$$ Now notice that $DA$ is the $D-\text{Symmedian}$ of $\triangle EDF$. Also $\angle QDC=\angle ACB=\angle ADB\implies AD$ and $DQ$ are isogonal. So, $Q$ must be the midpoint of $EF$. $\blacksquare$. Now just reverse these steps and get that $P$ is the midpoint of $EF$ using the fact that $Q$ is the midpoint of $EF$. (too lazy to type ).

I don’t know why amar_04 is nowadays posting so complicated solutions, all are going above my head Anyway angle chase to show that $APEB,APCF$ are both cyclic, so the result follows from the similarity of triangles $APB,APC$ with triangles $EQD,FQD$.