$PI*PD=PA*PF$ since $\angle FAI=90$ we have $\angle FDI=90$. also $FAE\sim FEP$ so $FE^2=FA*FP$ also $FE^2=FD^2+ED^2$ and $FP^2=PD^2+FD^2$ combining these ywe get $DQ^2=DI*DP$ and $\angle DQP=\angle DQI$ $M$ the midpoint of $BC$ let $PI$ intersect the circumcircle of $ABC$ at $K$ again $\angle DMQ=90=\angle DKQ$ so $DKQM$ is cyclic denote $\angle DQP=x$ now we have $\angle DKM=x$. now let the incircle touch $BC$ ar $S$. we will need $\angle MIS=\angle PIA=x$.This can be a useful lemma and it requires to prove $IMS\sim API$ but if we choose $N$ on the circumcircle such that $PN||AC$ and $L=QN\cap AC$ and $G$ on $AC$ such that $QG=QB$ and $G$ is different from $L$ we will easily get $NLC\sim AIT$ where incircle touches $AC$ at $T$ now notice $LC=CG/2=(AC-AB)/2=SM$ so the similarity we prove will give $r/SM=AI/AP$ since $AP=NC$ finally giving $IMS\sim IKQ$ and of course $ISK\sim IMQ$. let $U$ be the midpoint of $IM$ since $\angle ISU=\angle SIU=\angle IMP=180-\angle IMQ=180-\angle ISK$ we have that $S,K,U$ are collinear so $SK$ is the median in $IKM$ but $\angle SKM=x-\angle SKI=x-\angle IQP=\angle KPO=\angle OKP$ so $OK$ is the symedian in $IKM$ and $\angle OMK=x=\angle MKI$ so $MO$ is tangent to circle $IKM$ and that means (since $OK$ is the symedian in $IKM$) that $OI$ is also tangent to circle $IKM$ and $OI=OM$ now square this to get $R^2-2r*R=OI^2=OM^2=R^2-BM^2$ so $BM^2=2R*r$ dividing this by $R^2$ gives the result

A little typo in question, R is radius of the circumcircle and r is radius of the incircle, as always.