Hello xxravenxx Originally Posted by

**xxravenxx** ... Problem 12

Let M be a point inside the angle A. From point M perpendiculars MP and MQ are dropped to the angle sides. From point A, perpendicular AK is dropped to segment PQ. Prove that angle PAK = angle MAQ.

$\displaystyle \angle APM = \angle AQM = 90^o$ (given)

$\displaystyle \Rightarrow APMQ$ is a cyclic quadrilateral (opp angles supplementary)

$\displaystyle \Rightarrow \angle APQ = \angle AMQ$ (angles in same segment)

So in $\displaystyle \triangle$'s $\displaystyle APK, AMQ:$$\displaystyle \angle APK = \angle AMQ$ (proven)

$\displaystyle \angle AKP = \angle AQM$ ($\displaystyle =90^o$, given)

$\displaystyle \Rightarrow \angle PAK = \angle MAQ$ (angle sum of triangle)

Grandad