AoPS - Problem

Source: 2019 Saudi Arabia BMO TST III p2

Tags: geometry, trapezoid, right angle, equal segments

parmenides51

25.07.2020 11:37

Let $ABCD$ is a trapezoid with $\angle A = \angle B = 90^o$ and let $E$ is a point lying on side $CD$. Let the circle $\omega$ is inscribed to triangle $ABE$ and tangents sides $AB, AE$ and $BE$ at points $P, F$ and $K$ respectively. Let $KF$ intersects segments $BC$ and $AD$ at points $M$ and $N$ respectively, as well as $PM$ and $PN$ intersect $\omega$ at points $H$ and $T$ respectively. Prove that $PH = PT$.

Let $\ell$ be the reflection of line $AF$ over $FN$, and let $\ell$ intersect $AD$ at $G$, so $FN$ bisects $\angle AFG$. Then $\angle GFN = \angle AFN = \angle EFK = \angle EKF = \angle BKM$. Also, $\angle GNF = 180 - \angle ANF = \angle BMK$. Thus, $\triangle GFN \sim \triangle BKM$. Then $\frac{BP}{BM} = \frac{BK}{BM} = \frac{FG}{GN} = \frac{FA}{AN} = \frac{AP}{AN}$. Since $\angle PBM = \angle PAN =90$, it follows that $\triangle BPM \sim \triangle APN$. Then $\widehat{PH} = 2 \angle BPH = 2 \angle APT = \widehat{PT}$. Thus, $PH=PT$.