Problem

Source: Peruvian IMO TST, 2018

Tags: geometry, circumcircle



Let $ABC$ be, with $AC>AB$, an acute-angled triangle with circumcircle $\Gamma$ and $M$ the midpoint of side $BC$. Let $N$ be a point in the interior of $\bigtriangleup ABC$. Let $D$ and $E$ be the feet of the perpendiculars from $N$ to $AB$ and $AC$, respectively. Suppose that $DE\perp AM$. The circumcircle of $\bigtriangleup ADE$ meets $\Gamma$ at $L$ ($L\neq A$), lines $AL$ and $DE$ intersects at $K$ and line $AN$ meets $\Gamma$ at $F$ ($F\neq A$). Prove that if $N$ is the midpoint of the segment $AF$ then $KA=KF$.