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Let be a circumscribed circle to the rectangle $ABCD$ and $M$ a point which is on the circle. If $H_{1}$ is the orthocentre for $\triangle ABM$, $H_{2}$ for $\triangle BCM $, $H_{3}$ for $\triangle CDM$ and $H_{4}$ for $\triangle DAM$ prove that $H_{1}H_{3} \perp H_{2}H_{4}$. I will insert a picture for more details.

triangle thanks :)

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Let the foot of the perpendicular from $M$ to $AB, BC, CD, DA$ be $D_1, D_2, D_3, D_4$ respectively. The orthocenter $H_i$ lies on the line $MD_i$.

Since $AB, CD$ are parallel, $M, H_1, D_1, H_3, D_3$ are collinear and the line through the points is parallel to $BC$. Similarly, $M, H_2, D_2, H_4, D_4$ are collinear and the line through the points is parallel to $CD$. We have

$H_1H_3 \parallel BC \perp CD \parallel H_2H_4$

$\therefore H_1H_3\perp H_2H_4$

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