Let $PQRS$ be a quadrilateral that has an incircle and $PQ\neq QR$. Its incircle touches sides $PQ,QR,RS,$ and $SP$ at $A,B,C,$ and $D$, respectively. Line $RP$ intersects lines $BA$ and $BC$ at $T$ and $M$, respectively. Place point $N$ on line $TB$ such that $NM$ bisects $\angle TMB$. Lines $CN$ and $TM$ intersect at $K$, and lines $BK$ and $CD$ intersect at $H$. Prove that $\angle NMH=90^{\circ}$.
Problem
Source: 2022 Thailand MO Day 2 P9
Tags: geometry
MarkBcc168
16.06.2022 05:22
By Newton's theorem, $C,D,T$ are colinear. If $X=MR\cap BT$, then $(BT;NX)=-1$. Now, look at the pencil from $M$ and realize that you are done.
JAnatolGT_00
17.06.2022 23:44
By projective transformation $T\in CD\implies (MB,MT;MN,MH)=-1\implies MN\perp MH.$
Tintarn
28.06.2022 13:23
Here is a solution without projective geometry: As is well-known, $C,D,T$ are collinear (one can also use Pascal...). Now, we claim that $MH$ is the angular bisector of $\angle TMC$. This will clearly finish the proof, as then $\angle NMH$ is half of the $180^\circ$ angle $\angle BMC$. But this fairly general claim can now be checked by Ceva in the triangle $BTC$.