AK_|_ BC wanted, 2 circles related

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AK_|_ BC wanted, 2 circles related

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Source: 2020 Korea Junior Math Olympiad p2 KJMO

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#1 h Nov 24, 2021, 7:42 PM

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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 $BC$ .

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#2 h Nov 25, 2021, 4:29 AM • 2 Y

Y by geometry6, Hamzaachak

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.

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#3 h Nov 25, 2021, 5:20 PM • 1 Y

Y by teomihai

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 $K$ .

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#4 h Jun 18, 2023, 5:58 PM

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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.

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