Let $AP$ and $BQ$ be the altitudes of acute-angled triangle $ABC$. Using a compass and a ruler, construct a point $M$ on side $AB$ such that $\angle AQM = \angle BPM$.
parmenides51 wrote:
Let $AP$ and $BQ$ be the altitudes of acute-angled triangle $ABC$. Using a compass and a ruler, construct a point $M$ on side $AB$ such that $\angle AQM = \angle BPM$.
Choose $M$ such that $(MPQ)$ is tangential to $AB$. We claim this is our required point. Indeeed note that $$\angle BPM = 180^{\circ} - \angle BAC - \angle MPQ = 180^{\circ} - \angle BAC - \angle AMQ = \angle AQM \> \blacksquare$$