Problem 2285

Since $\angle PAC = \angle BAQ$ the right-angle triangles $PAP_B$ and $QAQ_C$ have equal angles and hence are similar. It follows $\frac{|AP|}{|AQ|} = \frac{|AP_B|}{|AQ_C|}$. We also have $\angle BAP = \angle BAC - \angle PAC = \angle BAC - \angle BAQ = \angle QAC$. Hence the right-angle triangles $PAP_C$ and $QAQ_B$ also have equal angles and are similar. It follows $\frac{|AP|}{|AQ|} = \frac{|AP_C|}{|AQ_B|}$. Combining both ratios we get
$$|AP_B| \cdot |AQ_B| = |AP_C| \cdot |AQ_C| \text{ oziroma } A\vec{P}_B \cdot A\vec{Q}_B = A\vec{P}_C \cdot A\vec{Q}_C,$$
since the points $P_B, Q_B, P_C$, and $Q_C$ all lie on the sides of the triangle. By the Power of a Point Theorem the points $P_B, Q_B, P_C$, and $Q_C$ are concyclic. Denote their common circle by $\mathcal{K}_1$.

Similarly, since $\angle CBP = \angle QBA$ the right-angle triangles $PBP_A$ and $QBQ_C$ are similar. Since $\angle ABP = \angle QBC$ the right-angle triangles $PBP_C$ and $QBQ_A$ are also similar. It follows $\frac{|BP_A|}{|BQ_C|} = \frac{|BP_C|}{|BQ_A|}$. Therefore
$$B\vec{P}_A \cdot B\vec{Q}_A = B\vec{P}_C \cdot B\vec{Q}_C,$$

Let $M$ be the midpoint of the segment $PQ$. The center of the circle $\mathcal{K}_1$ lies on the intersection of bisectors of the segments $P_BQ_B$ and $P_CQ_C$. Since both bisectors pass through $M$, $M$ is the center of the circle $\mathcal{K}_1$. Similarly, the center of the circle $\mathcal{K}_2$ lies on the intersection of bisectors of the segments $P_AQ_A$ and $P_CQ_C$. These bisectors also pass through $M$, hence $M$ is the center of the circle $\mathcal{K}_2$. Therefore the circles $\mathcal{K}_1$ and $\mathcal{K}_2$ have a common center, and both pass through the points $P_C$ and $Q_C$, hence they are the same. It follows that the points $P_A, P_B, P_C, Q_A, Q_B$, and $Q_C$ all lie on a common circle.