The problem was at this deleted question originally.
Given:
1) $\triangle ABC$ -- an arbitrary triangle
2) with circumcircle $\omega$ centered at $O$
3) and incenter $I$.
4) Let $D$ be the second intersection $AI$ with $\omega$,
5) $P$ be the point of intersection of line $BC$ and the line, perpendicular to $AI$, passing through $I$,
6) Let $Q$ be the point, symmetrical of $I$ through $O$.Prove that $\angle PAQ=\angle PDQ=90^\circ$
Disclaimer: this is not in a way my solution, I simply asked a person with strong contest math background to look at the problem, they gave me the solution then. I don't think I can solve geometry problems such a way.
I) 1) Let the continuation of $OD$ intersects $\omega$ in $D'$.
2) $IO=OQ$ and $DO=OD'=R$ $\Rightarrow$ $DID'Q$ -- parallelogram $\Rightarrow$
3) $D'I||QD$.
II) 4) Let $X=PD\cap\omega$,
5) let $\angle A=2\alpha$;
6) let $\angle B=2\beta$ $\Rightarrow$
7) $\angle IBC=\beta$ as $CI$ is bisector of $\angle ABC$
8) $ID=DC$ by the trillium theorem $\Rightarrow$
9) $\angle CID=90^\circ-\beta$ because
10) $\angle IDC=2\beta$ as an inscribed angle, subtended by the same arc $AC$, as $\angle ABC$ $\Rightarrow$
11) $\angle PIC=\beta$ $\Rightarrow$
12) $\triangle ICP\sim \triangle BIP$ $\Rightarrow$
13) $\displaystyle\frac{IP}{BP}=\frac{CP}{IP}$ $\Rightarrow$
5'-13') And even a faster way: $DC=DI=DB$ by the trillium theorem, so $D$ is center of circle $BIC$, but $IP\perp ID$ and $ID$ being a radius of circle $BIC$ makes $IP$ tangent to circle $BIC$ thus $IP^2=CP\cdot BP$ by the secant-tangent theorem,
14) $IP^2=CP\cdot BP$ -- is the power of point $P$ relative to $\omega$ $\Rightarrow$
15) $IP^2=PX\cdot PD$ $\Rightarrow$
16) $\triangle PXI\sim\triangle PID$ $\Rightarrow$
17) $D',I,X$ are collinear because $DD'$ is a diameter (expanded: $\angle PXI=\angle PID$ from similarity $\Rightarrow$ $\angle DXI$ $=180^\circ-\angle PXI$ $=180^\circ-90^\circ$ $=90^\circ$, but $\angle DXD'$ is subtended by diameter $DD'$ $\Rightarrow$ $\angle DXD'=90^\circ$ $\Rightarrow$ $\angle IXD'$ $=\angle IXD-\angle DXD'=0$)
III) 18) $D'I||QD$; $\angle D'XP=90^\circ$ $\Rightarrow$
19) $\angle QDP=90^\circ$
20) Center of the circle $QDP$ is the middle of $QP$.
21) $QO=OI$ and if we pass a line through $O$ which is $||IP$,
i.e. perpendicular to $AD$ then it will pass through center of circle $QDP$ (we can use converse Thales' theorem to see it)
22) but the line is bissector perpendicular of $AD$ $\Rightarrow$
23) $A$ lies on the circle $QDP$ (indeed, the line connecting the centers of two circles and thus containing their diameters is perpendicular to a chord $AD$ of $\omega$ and perpendicular to a chord of circle $QDP$, containing $D$, so the chords are the same) $\Rightarrow$
24) $\angle QAP=180^\circ-\angle QDP=90^\circ$