Denote $E \equiv AM \cap BC.$ It is well-known that the midpoint $M$ of arc $\widehat{BAC}$ is the circumcenter of $\triangle BIC.$ Thus, the inversion $\mathcal{I}$ with center $M$ and power $MI^2$ fixes $B, I, C.$ Hence, $\mathcal{I}$ swaps $\Gamma$ and $BC$, implying that $\mathcal{I}$ swaps $D, P$ and $E, A.$ Therefore, $\angle MPA = \angle MED$ and $\angle MPI = \angle MID.$ Hence, \[\angle API = \angle MPA - \angle MPI = \angle MED - \angle MID = \angle IDE = 90^{\circ}.\]

Let the incircle touch the side $AB$ at $E$. Since $M$ is the midpoint of the arc $\widehat{BC}$ not containing $A$ we have $\angle BPD = \angle MPC$ and $\angle PBD = \angle PMC$ so $P$ is the center of the spiral similarity sending $B$ to $M$ and $D$ to $C$. Moreover $\angle EBD = \angle IMC$ so since triangle $EBD$ and $IMC$ are both isosceles triangles, this spiral similarity sends $E$ to $I$ aswell. This implies that $\angle EPI = \angle BPM = \angle EAI$ so $APEI$ is cyclic, hence $\angle API = \angle AEI = 90^{\circ}$.

Dear Mathlinkers, have a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=504539 Sincerely Jean-Louis

My solution: By $p_{C}(X)$ we will denote the power of point $X$ WRT circle $C$.We have $$PD\cdot MD=p_{\circ(ABC)}(D)=BD\cdot CD.$$It's well known that $M$ is the circumcenter of $\triangle{IBC}$,so $$BD\cdot CD=p_{\circ(BIC)}(D)=MI^2-DM^2.$$From the above two relations we deduce that $PD\cdot MD=MI^2-DM^2$.This is equivalent to $MD\cdot MP=MI^2$,implying that $\triangle{PIM}\sim\triangle{IDM}$. Thus,$m(\angle{MPI})=m(\angle{DIM})=m(\angle{ABC})+\frac{m(\angle{BAC})}{2}-90^0$. Now,let $Q\equiv PI\cap\circ(ABC)$. Then,$$m(\angle{BAQ})=m(\angle{BAM})+m(\angle{MAQ})=m(\angle{BAM})+m(\angle{MPQ})=$$$$=m(\angle{BAM})+m(\angle{MID})=\frac{m(\angle{BAC})}{2}+m(\angle{ABC})+\frac{m(\angle{BAC})}{2}-90^0=$$$$=90^0-m(\angle{ACB}).$$But $m(\angle{BQA})=m(\angle{ACB})$,so $m(\angle{ABQ})=90^0$. Therefore $m(\angle{API})=m(\angle{APQ})=m(\angle{ABQ})=90^0$.

Let the incircle touches $AB$ at $X$ let the circumcircle of $AIX$ touches the circumcircle of $ABC$ at $p$ we want to show that $p,D,M$ are colinear let $AP$ cuts $BC$ at $J$ by the kown lemma $JI$ is perpendicular to $AM$ we khow angles $PCD=PAX=PIX$ we know that $PAI=PIJ=PDJ$ so triangles $PDC,PXI$ are similar so angles $DPX=XPI=XAI$ so we are done

I have a different solution posted here months ago. https://www.facebook.com/photo.php?fbid=10201117460374813&set=p.10201117460374813&type=3&theater

Attachments:

Inversion wrt $(BIC)$ sends $(AIP)$ to $(IND)$ , orthogonal to line $(AI)$ . Hence $(AIP)$ is orthogonal to $(AI)$ . And the result follows .

Let $E$, $F$ be tangency points of ($I$) with $CA$, $AB$ Since: $P$ $\equiv$ $DM$ $\cap$ ($\Gamma$) ($P$ $\ne$ $M$), we have: $PD$ is internal bisector of $\widehat{BPC}$ So: $\dfrac{PB}{PC}$ = $\dfrac{BD}{CD}$ = $\dfrac{BF}{CE}$ But: $\widehat{PBA}$ = $\widehat{PCA}$ then: $\triangle$ $PBF$ $\sim$ $\triangle$ $PCE$ Hence: $\widehat{BPF}$ = $\widehat{CPE}$ or $\widehat{EPF}$ = $\widehat{BPC}$ = $\widehat{BAC}$ So: $P$ $\in$ ($AEF$) or $\widehat{API}$ = $90^o$