For a scalene triangle $ABC$ with an incenter $I$, let its incircle meets the sides $BC, CA, AB$ at $D, E, F$, respectively. Denote by $P$ the intersection of the lines $AI$ and $DF$, and $Q$ the intersection of the lines $BI$ and $EF$. Prove that $\overline{PQ}=\overline{CD}$.
Problem
Source: KMO 2022 P5
Tags: geometry, incenter
Dlavez
29.10.2022 15:52
It is well-known that $\angle APC = \angle BQC = 90 ^{\circ}$. Notice that $\overline{CQ} \parallel \overline{DF}$ and points $D$, $P$, $C$, $Q$, $E$, $I$ lies on the circle with diameter $CI$. This leads to that $\square CPDQ$ is isosceles trapezoid, which finishes the problem. $\blacksquare$
China_BW
03.11.2022 14:31
Dlavez wrote: It is well-known that $\angle APC = \angle BQC = 90 ^{\circ}$. Notice that $\overline{CQ} \parallel \overline{DF}$ and points $D$, $P$, $C$, $Q$, $E$, $I$ lies on the circle with diameter $CI$. This leads to that $\square CPDQ$ is isosceles trapezoid, which finishes the problem. $\blacksquare$ Absolutely right
IAmTheHazard
10.02.2023 00:07
ew c indexed geo It is well-known that $\angle APC=\angle BQC=90^\circ$, hence $CDEPQ$ is cyclic. Thus it suffices to show that $\angle PIQ=180^\circ-\angle CID$, which is true because both are equal to $90^\circ+\frac{\angle C}{2}$. $\blacksquare$