Balti Tee 1990 · Ülesanne 8

Solution:

Let $O$ be the circumcentre of the triangle $ABC$ and $\angle B$ be its maximal angle (so that $\angle A$ and $\angle C$ are necessarily acute). Further, let $B_{1}$ and $C_{1}$ be the base points of the perpendiculars drawn from the point $P$ to the sides $AC$ and $AB$ respectively and let $\alpha$ be the angle between the Simson line $l$ of point $P$ and the height $h$ of the triangle drawn to the side $AC$. It is sufficient to prove that $\alpha=\frac{1}{2} \angle POB$. To show this, first note that the points $P, C_{1}, B_{1}, A$ all belong to a certain circle. Now we have to consider several sub-cases depending on the order of these points on that circle and the location of point $P$ on the circumcircle of triangle $ABC$. Figure 2 shows one of these cases - here we have $\alpha=\angle PB_{1}C_{1}=\angle PB_{1}C_{1}=\angle PAB=\frac{1}{2} \angle POB$. The other cases can be treated in a similar manner.

Figure 2