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

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? • As a beginner in Olympiad math just seeing 5+ circles in one diagram makes me anxious Aug 21, 2020 at 17:19
• @l1mbo lol, even I was like you , but now, the more the circles in the diagram , more I like the problem :) Aug 22, 2020 at 0:21
• Have you guys learned calculus, linear algebra and simillar topics?
– 1b3b
Aug 22, 2020 at 0:31
• @1b3b no , not in school atleast and I know only basic calculus.. Aug 22, 2020 at 1:18
• @Shubhangi, yes. I am interested in what math competitors learn in other countries. For example, I qualified for the state (Croatia) competition this year, but I learned Calculus 1 and basics of Calculus 2 on my own this summer (when I said learned I meant on understanding all rules, theorems, etc.). Also a lot of other topics are not even mentioned before college. But I think olympiad approach is better because it develops problem solving skills which is ground for math research
– 1b3b
Aug 22, 2020 at 11:03

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. $$\blacksquare$$