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Let $ABC$ be a triangle with orthocentre $H$ and circumcircle $\odot(ABC)$. Suppose $P\in\odot(ABC)$. Let $\gamma$ be Simson's line of $P$ wrt $ABC$.

Prove that $\gamma$ bisects $PH$.

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Simpson's Line The preceding link redirects to Wolfram Math World. –  Grobber Oct 28 '12 at 15:00
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Let $P_A,P_B,P_C$ be the projections of $P$ on the sides $BC,AC,AB$; let $Q$ be the intersection between $AH$ and the Simson's line of $P$. We want to show that $HQ=PP_A$. If we call $H_A$ the symmetric of $H$ with respect to the $BC$ side, we have that $H_A$ lies on the circumcircle of $ABC$, so, in order to prove that $HQ=PP_A$, it is sufficient to show that $PP_A QH_A$ is an isosceles trapezoid. Let $\theta=\widehat{P_C P_A B}=\widehat{P_B P_A C}$. Since $P P_A P_C B$ and $P P_A C P_B$ are cyclic quadrilaterals, we have $\theta=\widehat{P_C P B}=\widehat{P_B P C}$, too.

Now: $$\widehat{P_A Q H_A}=\frac{\pi}{2}-\theta,$$ and: $$\widehat{P H_A Q}=\widehat{P H_A A}=\widehat{P B A}=\widehat{P B P_C}=\frac{\pi}{2}-\theta.$$ Since $PP_A$ and $QH_A$ are both ortogonal to $BC$, $PP_A Q H_A$ is an isosceles trapezoid and we have $PP_A=QH$, QED.

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