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Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. $H$ is the foot of the perpendicular from $A$ to the line $BC$, and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $AB$ and $AC$, respectively.

Given that, $$AH^2=2\cdot AO^2$$ prove that the points $O,P,Q$ are collinear.

kvardekkvar
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Nadun Rajapaksha
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  • Doesn't this given mean that triangle $AHO$ is a right isosceless triangle at $O$? If you please upload a figure to your question, this will help you get more answers – Fareed Abi Farraj May 02 '19 at 19:39

1 Answers1

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Do an inversion centered at $A$ with radius $AH$. The image of $P$ is $B$ and the image of $Q$ is $C$. Denote by $O'$ the image of $O$ under that inversion. Since $AH^2 = AO \cdot 2AO$, we get $AO' = 2AO$, that is, $AO'$ is the diameter of the circumcircle of $ABC$. Since $A, B, C$ and $O'$ lie on a circle passing through the center of inversion, the images of $B, C$ and $O'$ lie on a line. These images are $P, Q$ and $O$.

kvardekkvar
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