Balti Tee 2019 · Ülesanne 11

Since $\{ACMP\}$ is cyclic, we have $\angle RPQ = \angle ACM$. Moreover, $\angle PQR = 90^\circ = \angle CMA$. Hence $\triangle PQR \sim \triangle CMA$. It follows that

$$\frac{PR}{PQ} = \frac{CA}{CM}. \qquad (1)$$

Similarly, $\angle PMB = \angle PAC$ and $\angle BPM = 90^\circ = \angle CPA$. Hence $\triangle BPM \sim \triangle CPA$. It follows that

$$\frac{CA}{BM} = \frac{CP}{BP}. \qquad (2)$$

Using (1), (2), and the equality $BM = CM$ we obtain

$$\frac{PR}{PQ} = \frac{CA}{CM} = \frac{CA}{BM} = \frac{CP}{BP}. \qquad (3)$$

Using (3) and $\angle QPB = 90^\circ = \angle RPC$ we obtain $\triangle QPB \sim \triangle RPC$. In particular, $\angle RCP = \angle PBQ$. Hence

$$\angle RCP = \angle PBQ = \angle CBA - \angle MBP = \angle ACB - \angle ACP = \angle PCB.$$