INAMO 2014 Shortlist G4 wrote: Given an acute triangle $ABC$ with $AB <AC$. Points $P$ and $Q$ lie on the angle bisector of $\angle BAC$ so that $BP$ and $CQ$ are perpendicular on that angle bisector. Suppose that point $E, F$ lie respectively at sides $AB$ and $AC$ respectively, in such a way that $AEPF$ is a kite. Prove that the lines $BC, PF$, and $QE$ intersect at one point. Solution:- We will restate the problem like this. Restated Problem wrote: Given an acute triangle $ABC$ with $AB <AC$. Points $P$ and $Q$ lie on the angle bisector of $\angle BAC$ so that $BP$ and $CQ$ are perpendicular on that angle bisector. Let $F$ be any arbitary point on $AC$. Now $FP$ intersects $BC$ at a point $N$ and $QN$ intersects $AB$ at a point $E'$. Then prove that $AE'PF$ is a kite. Note that this is the same problem as above. Let $BP\cap AC=Y$ and $BQ\cap AC=X$. Now notice that due to congruence $ABQY$ is a kite. Hence, $QB=QY$. So, $\angle YQC=\angle BYQ=\angle YBQ=\angle CQX$ also $\angle AQC=90^\circ$. Hence, $(A,C;Y,X)$ is harmonic. So, $$-1=(A,C;Y;X)\overset{B}{=}(E',N;QE'\cap BY,Q)$$Now notice that $\angle BPQ=90^\circ$. Hence, $\angle E'PB=\angle BPN=\angle FPY$. Now again notice the Congruence between triangles $E'PB$ and $FPY$. Hence, $E'B=FY$ also $AB=AY$. Hence, $AE'=AF$ also we got earlier that $E'P=PF$. Hence, $AE'PF$ is a kite. $\blacksquare$. @below thanks, fixed it.

amar_04 wrote: Now notice that $\angle BPQ=90^\circ$. Hence, $\angle E'PB=\angle BPQ=\angle FPY$. A typo there, $\angle E'PB=\angle BPN $