Claim: M is on BC angle ABP = 180° - angle BMP angle ACP = 180° - angle CMP 180° = angle ABP + angle ACP = 360° - (angle BMP + angle CMP) Thus, angle BMP + angle CMP = 180° -> B, M, C are collinear. Claim: AB = BP angle ABD = angle BMD = angle BPD Hence, triangle ABO is congruent to triangle BPO AB = BP Hence, angle MCP = angle BCP = angle ACB = angle ACM = angle MPC Thus, we get MP = MC (proven).

$\angle BMP=180-\angle PBA=\angle PCA=180-\angle PMC$, so $B, M, C$ are collinear. $\angle MCP\equiv\angle BCP=\frac{1}{2}\angle BOP=\frac{1}{2}(180-2\angle OPB)=\frac{1}{2}(180-2\angle OBA)=\frac{1}{2}\angle AOB=\angle ACB\equiv\angle ACM=\angle MPC$, so $MP=MC$.

We have $\angle{OPB} = \angle{OPB} = \angle{OBA} = \angle{OAB}$. Then $\angle{AOB} = \angle{BOP}$. From this, we have $\angle{MPC} = \angle{ACB} = \dfrac{1}{2} \angle{AOB} = \dfrac{1}{2} \angle{BOP} = \angle{BAP} = \angle{BCP}$. So $\triangle MCP$ is $M$ - isosceles or $MC = MP$

Observe that $$\angle PMC=\pi-\angle PCA=\angle ABP=\angle ABO+\angle OBP=\pi-\angle BOP=\pi-\angle BMP$$Thus, $B$, $M$ and $P$ are collinear. On the other hand, $$\angle OCM=\angle OBM=\angle OPM$$which implies $MP=MC$ as desired.