Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^\circ$.

Question

In convex cyclic quadrilateral $ABCD$, we know that lines $AC$ and $BD$ intersect at $E$, lines $AB$ and $CD$ intersect at $F$, and lines $BC$ and $DA$ intersect at $G$. Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^\circ$.

My Progress:

Claim : $PBQD$ is cyclic

Proof: Note that $CQ\cdot CD=CE\cdot CA=CB\cdot CP \implies PQDB$ is cyclic.

Claim: $APQC$ is cyclic

Proof : angle chase! Note that for this to be true , it is enough to show that $\angle AEB=\angle AQC$ or it is enough to show that $\angle AEB=\angle AQC $ or it is enough to show that $\angle AED=\angle AQD$ which is true since $AEDQ$ is cyclic.

Claim: $E\in PQ$

Proof: So enough to show that $\angle AEQ+\angle AEP=180 $

or enough to show that $180- \angle ADC + \angle AEP=180 $

or enough to show that $\angle ADC= \angle ABC$ , which is true since $ABCD$ is cyclic.

after that I am stuck.

I observed that $FG , AM, PQ$ concur but was not able to prove. Can someone give hints?

Thanks in advance.

Answer 1

Continuing from where you finished...

So, we have $PBDQ$ cyclic and $E\in PQ$. Now focus on quadrilateral $PBDQ$. From definition $A$ is the Miquel Point of the quadrilateral $PBDQ$. Now let $X:=PD\cap BQ$ and thus, by Miquel point properties, we get that $A$ is projection of $X$ on $CE$. Thus, its enough to show that $M,A,X$ are collinear but this is trivial. Just apply Pappus Theorem on $\{PGB,QFD\}$ completing the proof. $\blacksquare$