Baltic Way 2019

Since $\{ACMP\}$ is cyclic, we have $\angle RPQ=\angle ACM$. Moreover, $\angle PQR=90^{\circ }=\angle CMA$. Hence $△PQR\sim △CMA$. It follows that $$\frac{PR}{PQ}=\frac{CA}{CM}.(1)$$ Similarly, $\angle PMB=\angle PAC$ and $\angle BPM=90^{\circ }=\angle CPA$. Hence $△BPM\sim △CPA$. It follows that $$\frac{CA}{BM}=\frac{CP}{BP}.(2)$$ Using (1), (2), and the equality $BM=CM$ we obtain $$\frac{PR}{PQ}=\frac{CA}{CM}=\frac{CA}{BM}=\frac{CP}{BP}.(3)$$ Using (3) and $\angle QPB=90^{\circ }=\angle RPC$ we obtain $△QPB\sim △RPC$. In particular, $\angle RCP=\angle PBQ$. Hence $$\angle RCP=\angle PBQ=\angle CBA-\angle MBP=\angle ACB-\angle ACP=\angle PCB.$$