I'm given the following problem and hint, but I solved the problem another way. I was wondering whether my proof is correct; namely, am I making any wrong assumptions? Thank you.

Altitude $AD$ of $\triangle ABC$ has been extended to meet the circumcircle at point $X$. Point $H$ was chosen on line $AD$ so that $HD=DX$. Show that $BH$ is perpendicular to $AC$. HINT: Let $P$ be the point where $BH$ crosses the circle and show that arc $PAC$ is equal in measure to arc $XC$.

Proof. Since $HD=DX$ and H lies on the altitude AD, we can see that X is the reflection of H across the line BC. In particular, the circumcircle of $\triangle BHC$ is the locus of all points whose reflection on line BC lie on the circumcircle of $\triangle ABC$. It suffices, then, that $CA_{1}\perp BX$ when $BH\perp AC$. Notice that $\angle ABC=\angle A_{1}BC$. Then, $$90^{\circ}=\angle ADB\newcommand{arc}[1]{\stackrel{\circ}{#1}}\arc{=}\dfrac{1}{2}(\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}+\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{CX})\newcommand{arc}[1]{\stackrel{\circ}{#1}}\arc{=}\dfrac{1}{2}(\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{BY}+\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{CX})=\angle CUX,$$as desired.

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Your proof looks entirely correct to me. And I like the nice simplification you use to get there.

A typo:

... whose reflection on line BC lie on the circumcenter of $\triangle ABC$.

should read

... whose reflection on line BC lie on the $\color{red}{\text{circumcircle}}$ of $\triangle ABC$.

  • 1
    $\begingroup$ Fixed thank you! $\endgroup$ – Alex D Apr 9 '18 at 13:34

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