The acute triangle $ABC$ satisfies $\overline {AB}<\overline {BC}<\overline {CA}$. Denote the foot of perpendicular from $A,B,C$ to opposing sides as $D,E,F$. Let $P$ a foot of perpendicular from $F$ to $DE$, and $Q(\neq F)$ a intersection point of line $FP$ and circumcircle of $BDF$. Prove that $\angle PBQ=\angle PAD$.
Problem
Source: 2021 Korea Winter Program Test2 Day2 #6
Tags: geometry
geo0420
17.02.2021 11:05
Let $PF \cap AD = R$. Since $\angle APF = \angle BPF$, it becomes just a simple angle chasing: \begin{align*} \angle PBQ &= \angle FQB - \angle BPF \\ &= \angle FDB - \angle APF \\ &= \angle A - \angle APF \\ &= 90^{\circ} - \angle ADE - \angle APF \\ &= \angle PRD - \angle APF \\ &= \angle PAD \;\; \blacksquare \end{align*}
Mogmog8
04.03.2022 08:18
[asy][asy] size(7cm); import geometry; pair A,B,C,D,E,F,P,Q; A=(-5.5,9); B=(-8,0); C=(4,0); D=foot(A,B,C); E=foot(B,A,C); F=foot(C,A,B); P=foot(F,D,E); Q=intersectionpoints(segment(P,F),circle(F,B,D))[0]; draw(A--D, grey); draw(B--E, grey); draw(C--F, grey); draw(E--D); draw(F--P); draw(A--P--B); draw(F--D); draw(A--B--C--cycle); draw(circle(B,F,D)); dot(A^^B^^C^^D^^E^^F^^P^^Q); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,SE); label("$E$",E,NE); label("$F$",F,NW); label("$P$",P,SE); label("$Q$",Q,SW); [/asy][/asy] Notice $(A,B;F,\overline{AB}\cap\overline{DE})=-1$ so $\overline{PF}$ bisects $\angle APB$ by the Right Angles and Bisectors Lemma. Also, \begin{align*}\measuredangle BQP&=\measuredangle BQF=\measuredangle BDF=\measuredangle CAB\\&=90-\measuredangle ABE=90-\measuredangle ADE=\measuredangle(FP,AD)\end{align*}so $\measuredangle PBQ=\measuredangle DAP.$ $\square$