$2010$ $G1$, proving quadrilateral concyclic

Question

I was solving the following question, from $2010\text{ IMO}$ shortlist.

$G1.$ Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$. The lines $BP$ and $DF$ meet at point $Q$. Prove that $AP=AQ$. ($L$ is $A$, $K$ is $B$, $P$ is $F$, and $Q$ is $P$)

While making the diagram on GeoGebra, I observed that $LQRP$, or $APQF$ in the question's labelling, is cyclic. I tried proving it, and I proved it in my mind. Then, as is characteristic of me, I forgot how I proved it. The only thing I remember is using $\angle LQP=\angle LMP=\angle LNP$, then saying 'Wow, that was easy.'

I've tried to do that again now, but I've run into a problem. We have that $\angle LQP=\angle LNP$. If $LQRP$ is concyclic, then we must have $\angle LRP=\angle LQP$ as well. So we have $\angle LNP=\angle LRP$, which is absurd. However, it is obvious from the diagram that the quadrilateral indeed is concyclic.

Where did I make the mistake? And how can I prove that it is concyclic?

Answer 1

$\angle LQP$ is NOT equal to $\angle LMP$. This would imply $LQMP$ is cyclic, but that would mean $P$ lies on the circumcircle of $LKM$ (the circle of $LQM$) but this is not true, as $P$ is the foot of an altitude.

Also note that $\angle LQP$ is not equal to $LNP$, as that would imply $LQNP$ is cyclic, but you can see from your diagram that isn't true.

Finally, observe that proving $LQRP$ is cyclic would solve the original problem.

You can show this with inversion, or some synthetic length computing with that idea; it isn't too difficult..