For convenience, rename $I \equiv O$ and let $D,E$ be the midpoints of $BC,BA.$ It's well known that $AI,MN,DE$ and the perpendicular dropped from $B$ to $AI$ concur at $P.$ For a short proof you may see the topic Another unlikely concurrence here. Hence, it follows that $AQ \perp BI$ and $PE \parallel CA$ implies that $\triangle CMN \sim \triangle DNP.$ Thus $\frac{MP}{MN}=\frac{CD}{CN}=\frac{BC}{2CN} \Longrightarrow \frac{MP}{BC}=\frac{MN}{2CN}$ But if $T \equiv MN \cap CI,$ we have $ TN=\frac{_1}{^2}MN$ $\Longrightarrow \frac{MP}{BC}=\frac{TN}{CN}.$ Since $\angle QAI= 90^{\circ}-\frac{_1}{^2}(\angle A+\angle B)=\angle TCN,$ we get $\triangle CNT \sim \triangle AIQ$ $\Longrightarrow \frac{IQ}{IA}=\frac{TN}{CN}=\frac{MP}{BC} \Longrightarrow MP \cdot IA=BC \cdot IQ.$

Correct me if I'm wrong, for convenience also, let I be the incenter. <AMI=90 so <IMC=90 also<INC=90 so MCNI is cyclic so <IMN=<INM=<C/2. Then in triangle APM we obtain that <APM=180-(<A/2+90+<C/2)=B/2, similarly in BQN we obtain that <BQN=<BQP=<A/2 . Then, quadrilateral ABPQ is cyclic, so BIP ~ AIQ so, IQ/AI=PI/BI, but we have to prove IQ/IA=MP/BC so it is enough to prove PI/BI=MP/BC, but this is true because triangles IMP ~ IBC (<IMP=<ICB=<C/2 and <IPM=<IBC=<B/2) so MP/BC=PI/BI so we're done

PP. Let $\triangle ABC$ with incircle $w=C(I,r)$ and $\begin{array}{ccc} M\in AC\cap w & ; & N\in BC\cap w\\\\ P\in AI\cap MN & ; & Q\in BI\cap MN\end{array}$ . Prove that $\left\{\begin{array}{c} MP\cdot IA=BC\cdot IQ\\\\ NQ\cdot IB=AC\cdot IP\end{array}\right\|$ . Proof. From an well-known property $\left\{\begin{array}{c} QA\perp QB\\\\ PA\perp PB\end{array}\right\|$ obtain that $AQPB$ is inscribed in the circle $\theta$ with diameter $[AB]$ . Thus, $\left\{\begin{array}{ccc} \triangle BIC\sim\triangle PIM & \implies & \frac {BC}{PM}=\frac {IB}{IP}\\\\ \triangle AIC\sim\triangle QIN & \implies & \frac {AC}{QN}=\frac {IA}{IQ}\\\\ p_{\theta}(I)=IA\cdot IP=IO\cdot IB & \implies & \frac {IB}{IP}=\frac {IA}{IQ}\end{array}\right\|$ $\implies$ $ \left\{\begin{array}{c} \frac {BC}{PM}=\frac {IA}{IQ}\\\\ \frac {AC}{QN}=\frac {IB}{IP}\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c} MP\cdot IA=BC\cdot IQ\\\\ NQ\cdot IB=AC\cdot IP\end{array}\right\|$ . See PP5 from here.

Let $\angle BAC= 2\alpha$, $\angle CBA=2\beta$ and, $\angle ACB=2\theta$. Notice that $\alpha +\beta+ \theta=90 (*)$, draw $ON$ and $OM$, these lines are perpendicular to $BC$ and $AC$ respectively, this means that $OMCN$ is a cyclic quadrilateral and since $OC$ is an angle bisector we get that $\angle OMN = \angle MNO = \theta$ and because of $(*)$ we see in triangles $\Delta BNQ$ and $\Delta AMP$ that $\angle OQN=\alpha$ and $\angle MPO= \beta$. Therefore $\Delta OMP \cong \Delta BOC$ and $\Delta ONQ \cong \Delta AOC$. Setting the relations for both pairs we get $\frac{OA}{OQ}=\frac{BC}{MP} \rightarrow MP \cdot OA=BC \cdot OQ$.

Dear Mathlinkers, I have decomposed this problem in 1. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=543815 2. https://artofproblemsolving.com/community/c6h1549498_two_perpendiculars 3. https://artofproblemsolving.com/community/c6t48f6h1549508_two_equal_angles 4. and finish with the similar triangles AIQ and BIP. Sincerely Jean-Louis