Source : Solution 3 of G6 in pg 71 of https://www.imo-official.org/problems/IMO2019SL.pdf
Problem : Let $I$ be the incentre of acute-angled triangle ABC. Let the incircle meet $BC, CA,$ and $AB$ at $D,E,$ and $F$, respectively. Let the line $EF$ intersect the circumcircle of the triangle at P and Q such that $F$ lies between $E$ and $P$. Prove that $\angle DPA+\angle AQD=\angle QIP$.
My questions : I understand everything until the last paragraph. How is the combination of "these two transformations" a projective map from the lines $AB,AC,AN,AU,AV$ to $IF,IE,IT,IP,IQ$? I'm also not sure why does this combination of maps preserve orthogonality when the solution says "On the other hand …"? Can you please help elaborate?
Using geogebra I tested the claim in the last paragraph to be true. That is: if $G$ is a point on the line PQ. Let $AG$ intersects $\Gamma$ (the circumcircle of ABC) again at $G_1$. Let $G_2$ be the inversion of $G_1$ wrt the circle centred at $R$ with radius $RB$. Then let $G'$ be the intersection of $AG_2$ with $PQ$. Then $AG\perp IG'$. I'm just not seeing why this is true from the last paragraph of the official soln.
Best Answer
I finally understood what is said in that last paragraph. Basically, this is true :
Let $a,b,c,d$ be concurrent lines, and $a',b',c',d'$ be also concurrent lines. Suppose there exists a projectivity between $(a,b,c,d)$ and $(a',b',c',d')$. If $\measuredangle(a,a')=\measuredangle(b,b')=\measuredangle(c,c')=\theta$, then
(1) $\measuredangle(d,d')=\theta$
(2) $\measuredangle(s,t)=\measuredangle(s',t')$ for all $s,t\in\{a,b,c,d\}$.