In an angle with vertex $A$, a point $M$ is chosen. Let $P$ and $Q$ be the feet of the perpendiculars from $M$ to the sides of the angle, and let $K$ be the foot of the perpendicular from $A$ to $PQ$. Prove that $\angle MAP = \angle QAK$.

1. Problem Description

In an angle with vertex $A$, a point $M$ is chosen. Let $P$ and $Q$ be the feet of the perpendiculars from $M$ to the sides of the angle, and let $K$ be the foot of the perpendicular from $A$ to $PQ$. Prove that $\angle MAP = \angle QAK$.

2. Solution Steps

Let's consider the quadrilateral $APMQ$. Since $\angle APM = 90^{\circ}$ and $\angle AQM = 90^{\circ}$, the quadrilateral $APMQ$ is cyclic with diameter $AM$. Let $O$ be the midpoint of $AM$. Then $O$ is the center of the circle passing through $A, P, M, Q$.

Since $APMQ$ is a cyclic quadrilateral, $\angle MAP = \angle MQP$.

Also, since $AK \perp PQ$, we have $\angle AKP = 90^{\circ}$.

Consider the triangle $APQ$. We have $\angle PAQ + \angle APQ + \angle AQP = 180^{\circ}$.

In right triangle $AKP$, $\angle PAK + \angle APQ = 90^{\circ}$.

In right triangle $AKQ$, $\angle QAK + \angle AQP = 90^{\circ}$.

Let $\angle MAP = \alpha$. Then $\angle MQP = \alpha$. We want to prove that $\angle QAK = \alpha$.

Since $\angle QAK + \angle AQP = 90^{\circ}$, we have $\angle QAK = 90^{\circ} - \angle AQP$.

If $\angle MQP = \alpha$, then $\angle AQP = \angle AQP$.

Also, $\angle PAK + \angle APQ = 90^{\circ}$, so $\angle PAK = 90^{\circ} - \angle APQ$.

Since $\angle PAQ + \angle APQ + \angle AQP = 180^{\circ}$, $\angle PAQ = 180^{\circ} - \angle APQ - \angle AQP$.

Now, $\angle PAQ = \angle PAM + \angle MAQ = \angle PAM + \angle MAP$.

Consider the cyclic quadrilateral $APMQ$.

$\angle PAM + \angle QMP = 180^{\circ}$ and $\angle QAM + \angle QPM = 180^{\circ}$.

We are given $AK \perp PQ$. In triangle $APQ$, let $\angle APQ = \beta$ and $\angle AQP = \gamma$. Then $\angle PAQ = 180^{\circ} - \beta - \gamma$.

Since $AK \perp PQ$, $\angle AKP = 90^{\circ}$. Then $\angle PAK = 90^{\circ} - \angle APQ = 90^{\circ} - \beta$.

Similarly, $\angle QAK = 90^{\circ} - \angle AQP = 90^{\circ} - \gamma$.

Then $\angle PAQ = \angle PAK + \angle QAK = (90^{\circ} - \beta) + (90^{\circ} - \gamma) = 180^{\circ} - \beta - \gamma$.

We want to prove $\angle MAP = \angle QAK$, which means $\angle MAP = 90^{\circ} - \gamma$.

Also, we know $\angle MAP = \angle MQP = \angle AQP = \gamma$.

Thus, we have $\gamma = 90^{\circ} - \gamma$, which means $2\gamma = 90^{\circ}$, so $\gamma = 45^{\circ}$. This is only true for specific values of the angles.

In triangle $APQ$, $\angle PAK + \angle QAK = \angle PAQ$.

Since $APMQ$ is cyclic, $\angle MAP = \angle MQP$.

We want to show $\angle QAK = \angle MAP = \angle MQP$.

Since $\angle QAK = 90 - \angle AQP$, we want $\angle AQP = \angle MAP$.

3. Final Answer

$\angle MAP = \angle QAK$