Let $ABC$ be an acute triangle with circumcircle $\Omega$ and $\overline{AB} < \overline{AC}$. The angle bisector of $A$ meets $\Omega$ again at $D$, and the line through $D$, perpendicular to $BC$ meets $\Omega$ again at $E$. The circle centered at $A$, passing through $E$ meets the line $DE$ again at $F$. Let $K$ be the circumcircle of triangle $ADF$. Prove that $AK$ is perpendicular to $BC$.
Problem
Source: 2020 Korea Junior Math Olympiad p2 KJMO
Tags: geometry, perpendicular
sarjinius
25.11.2021 07:29
Let $X$ be the midpoint of $EF$. Since $AE=AF$, $AX \perp EF$ and $AX \parallel BC$. To show that $AK \perp BC$, we need to show $AK \perp AX \iff AX$ is tangent to $(ADF)$. Let $Y$ be the reflection of $A$ across $X$. This implies that $Y$ is also the reflection of $A$ across $DE$, and since $DE$ is a diameter of $\Omega$, $Y$ must also lie on $\Omega$. By power of a point, $XA^2 = XA \cdot XY = XD \cdot XE = XD \cdot XF$. Therefore, $AX$ is tangent to $(ADF)$, and we are done.
Steve12345
25.11.2021 20:20
By taking the other intersection of the circle centered at $A$, passing through $E$ with the circumcircle, it's not hard to see that the reflection of $O$ over $AD$ is $K$.
rafigamath
18.06.2023 20:58
Let $\angle{ADF}=\alpha$, we will show that $\angle{KAD}=\alpha$. $\angle{ADF}=\alpha \implies \angle{AKF}=2\alpha$, which implies $\angle{KAF}=90-\alpha$, and since $ED$ is diameter, $\angle{DAF}=90-2\alpha$ and this implies the desired result.