Let ABC be an acute angled triangle with circumcenter O.

Kallie Arroyo 2022-06-02 Answered
Let ABC be an acute angled triangle with circumcenter O. A circle passing through A and O intersects AB, AC at P, Q respectively. Show that the orthocentre of triangle OPQ lies on the side BC.
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Step 1
First of all, let us call $\alpha =\mathrm{\angle }BAC$ and $\beta ,\gamma$ the other two. Using that APOQ is a cyclic quadrilateral, we have that $\mathrm{\angle }QPO=\mathrm{\angle }QAO=\frac{\pi }{2}-\beta$ , and similarly $\mathrm{\angle }PQO=\frac{\pi }{2}-\gamma$ .
Now, take the circunference through O,P,B, and let R be the intersection with the side $\overline{BC}$ . It is not so difficult to check that CQOR is a cyclic quadrilateral (and is an excellent excercise if you haven't done it before). This allows us to compute the angles
$\mathrm{\angle }RPO=\mathrm{\angle }OQR=\frac{\pi }{2}-\alpha .$ .
In particular, this shows that $\overline{PO}\perp \overline{RQ}$ and $\overline{QO}\perp \overline{PR}$ , showing that R is the orthocenter of PQO

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