The diameter \( AD \) of the circumcircle of triangle \( ABC \) intersects line \( BC \) at point \( K \). Point \( D \) is reflected symmetrically with respect to point \( K \), resulting in point \( L \). On line \( AB \), a point \( F \) is chosen such that \( FL \perp AC \). Prove that \( FK \perp AD \). Proposed by Matthew Kurskyi
Problem
Source: Kyiv City MO 2025 Round 1, Problem 10.3, 11.3
Tags: geometry
SBYT
20.01.2025 08:45
Let the line passes $K$ perpendicular to $AD$ meet $AC,AB$ at $E,F$,respectively. $\angle AEK=90^\circ -\angle DAC=\angle ABC$,so $E,F,B,C$ are concyclic. Then $EK\cdot KF=BK\cdot KC=AK\cdot KD$,so $E,F,A,D$ are concyclic. Recall that $AD\perp EF$,so $L$ is the orthocenter of $\bigtriangleup AEF$,then $FL\perp AC$.$\Box$.
Attachments:
ehuseyinyigit
20.01.2025 13:20
Let's prove the reverse of it, in other words: Reverse Problem: We take $F$ such that $FK\perp AC$. Let the perpendicular drawn from $F$ to $AC$ intersect $AD$ and $AC$ at $L$ and $H$, respectively. It is clear that $F$ is unique. We wish to prove $LK=KD$. Let $FK\cap AC$ be $E$. $BFKD$ is cyclic. Thus $\angle KFD=\angle KBD=\angle DAE$ gives $ADEF$ concyclic. $FK.KE=AK.KD=BK.KC$ implies $BCEF$ concyclic. Hence, $\angle FLD=\angle ADC=\angle ABC=\angle KDF$ yields us $LK=KD$ as desired.