Saudi Arabia Mathematical Competitions

Assume that $AB<AC$. We have $MB=MC=MI$. Indeed, from the hypothesis it follows $$\widehat{IBM}=\frac{1}{2}(\hat{A}+\hat{B})=\widehat{BIM},$$ hence triangle $BMI$ is isosceles.



Furthermore, since $$\widehat{BPM}=\frac{1}{2}\hat{A}=\widehat{MBC}$$ one has $△DBM\sim △BPM$, hence $M{B}^2=MD\cdot MP$. It follows $M{I}^2=MD\cdot MP$, so $△MDI\sim △MIP$, and consequently $$\begin{array}{c}\widehat{MPI}=\widehat{MID}.\end{array}\text{\hspace{1em}(1)}$$ Notice now that $ID\parallel OM$, implying $\widehat{MID}=\widehat{AMO}$. Let $N$ be the antipodal point of $M$ in circle $\mathcal{C}$. Finally, we have $$\begin{array}{c}\widehat{API}=\widehat{APM}-\widehat{MPI}=\widehat{APM}-\widehat{AMN}\\ =\frac{1}{2}(\text{overparen}AM-\text{overparen}AN)=\frac{1}{2}\text{overparen}MN=90^{\circ }\end{array}$$