Let $O$ be the centre of circle $AXY$.Also let $IK\cap BC=M$.$I$ is the incentre of $ABC$ part 1 $OYMX$ is cyclic with $\angle{OYD}=\angle{OXD}=90$ We have $\angle{YQI}=\angle{YAX}=1/2\angle{A}$ and $\angle{QIY}=180-\angle{BIC}=90-1/2\angle{A}$.So $\angle{QYI}=90$.Similarly $\angle{PXI}=90$.This implies $I$ to be the orthocentre of $PQK$.So the nine point circle of $PQK$ passes through $X,Y$, the midpt of $KI,$,the midpt of $PQ$.i.e,$O,Y,M,X$ are cyclic and it is well known that $\angle{OYM}=\angle{OXM}=90$ part 2 $A,O,D$ are collinear we have $\angle{AYP}=\angle{ADX}$.This is because since $Y$ is the incentre of $ABD$ and $DX$ is ext angle bisector of $\angle{ADB}$,$(A,Y,D,DX\cap BY)$ is cyclic.Now as $2\angle{AYP}=2\angle{ADX},\angle{AOP}=\angle{ADC}$ and $PQ\parallel BC(IK\perp BC,IK\perp PQ)$,$A,O,D$ are collinear part 3 $O,Y,D,M,X$ are concyclic We can easily see that $OYMX$ is harmonic quadrilateral.Also $DY,DO,DX,DM$ is harmonic pencil.hence $D$ must lie on circle $OYMX$ Rest is straightforward.We get $\angle{YMX}=\angle{YDX}=90$ so $\angle{YOX}=90$ and $\angle {A}=90$.Also,$\angle{ADM}=\angle{OYM}=90$

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Can you please explain part 3?

for a triangle I is fixed => k is fixed => perpendiculars l1 and l2 on BI and CI from k are fixed => l1 intersection CI (=Q') and l2 intersection BI (=P') are fixed which precisely has to be unique and equal to points Q and P. I have used the fact that QK is perpendicular CI and PK is perpendicular BI which is trivial angle chase. NOW THIS reduces the problem to show that if angle Q'AP'=90 => angle BAC=ADC=90 since points A,Q,Y,X,P are concyclic with angle QXP=90. take an incircle centred at origin whose touch points be 1,b,c. => A= 2a*b/(a+b), B=2b/(b+1), C=2c/(c+1), I=0, K=2. now Q' = 2c*x/(c+1) where x is some real number since Q' lies on line CI. KQ' is perpendicular to BI => (Q'-2)/B be purely imaginary => Q'=2c(b+1)/(c+b) => Q'-A = 2b/(b+c) { using z+cong(z)=0 iff z is img} similarly P'-A= 2c/(c+b) now since QA perpendicular to PA => (Q'-A)/(P'-A) = b/c is imaginary => B/C is imaginary => angle BAC=90. angle ADC=90 is trivial thereafter. This is how i killed the problem during test within mins xD

Solution - Let $J$ be the foot of perpendicular from $I$ on $BC$ , by given condition we have $J$ is midpoint of $IK$. Claim - $\square IXKY$ is cyclic with center $J$. Proof - $\angle IXK =\angle PIX+\angle XPI= \frac { B+C}2 + \angle XAY = \frac { B+C}2 + \frac A2 = 90 $ and similarly $\angle IYK = 90 $ , Hence proved. Now let $E,F$ be foot of perpendicular from $I$ on $AC$ and $AB$ , note that $IX\perp JE$ and $IJ=JX$ , this implies $X$ is reflection of $I$ in $JE$ and similar for $Y$ We will later prove a lemma showing that if the reflections of $I$ in these sides subtend angle $\frac{\angle BAC}2$ at the point $A$ then $\angle BAC=90$ As $DY$ and $DX$ are internal and ext angle bisectors we have $\angle XDY=90$ Moreover $\angle BAC=90$ gives $\angle XIY= 135$ which using claim gives $\angle XJY= 90 $ and $\angle YXJ= 45$ So $X,J,D,Y$ are concyclic and hence $\angle YDB= \angle YXJ =45$ but $DY$ is angle bisector of $\angle ADB$ Hence $\angle ADB=\angle ADC =90 $ Hence proved ....... ( except for the lemma of course ) For the lemma we rename the foot of perpendicular from $I$ on $BC$ as $D$ for convinience. Lemma - Let the $I$ be incenter of $\Delta ABC$ and let the incircle touch the sides at $D,E,F$ . Let $X,Y $ be reflections of $I$ in $DE$ and $DF$ respectively . If $\angle YAX = \frac{\angle BAC}2 $ ; then $\angle BAC= 90 $ Proof - Note that $\angle EDF = 90 - \frac{\angle BAC}2$ and hence the given condition is equivalent to $\angle EDF +\angle YAX=90$ We use COMPLEX NUMBERS Let incircle be the unit circle and let $D\leftrightarrow d , E\leftrightarrow e ,F\leftrightarrow f$ Then we have $X\leftrightarrow x = d+e , Y\leftrightarrow y= f+d$ and $A \leftrightarrow a= \frac {2fe}{f+e}$ Now the given condition says $$Arg ( \frac {e-d}{f-d} ) + Arg ( \frac { y-a}{x-a}) = \frac { \pi}2 $$$$\implies \frac {(e-d)(y-a)}{(f-d)(x-a)} \in \mathbb{R}i$$$$\implies \frac {(e-d)(y-a)}{(f-d)(x-a)} =-\frac {(\bar e-\bar d)(\bar y-\bar a)}{(\bar f-\bar d)(\bar x-\bar a)}$$$$\implies \frac {fd+de+f^2-ef}{fd+de+e^2-ef}=-\frac{fe+f^2+de-df}{fe+e^2+df-de}$$$$\implies (e+f)(e^2+f^2)=0$$Now the fact that $a$ is defined implies that $e+f\neq 0$ and hence $e^2+f^2 = 0 \implies e=\pm if$ and hence $\angle EIF= 90 $ which implies $\angle BAC =90$ Hence proved !!!