Find the angle between $AC$ and $IQ$ 
Incenter of triangle $ABC$ is point $I$. Points $M$ and $N$ are respectively middle points of $AB$ and $AC$. Intersection point of line $CI$ and $NM$ is $P$. There is such point $Q$, so that $MN$ and $PQ$ would be perpendicular, and $BI$ with $QN$ parallel to each other.
Find the angle between lines $AC$ and $IQ$.


What I did was to write down all the angles that we have - there are some useful conclusions  that got out of this (for example $PN=AN=NC$, therefore angle $APC$ is right), also since no ratios or angles are given in the problem, I have a slight suspicion that the answer is $90$ degrees.
 A: The answer is $90 ^{\circ}$

Let's denote the points of tangency of the inscribed circle with $BC, AC, AB$ by $T_a, T_b, T_c$.

Lemma. ("A-bisector, B-midline, C-touchchord") The point P lies on $T_cT_a$

This is a well-known fact and can be proven by mass point geometry. It will come in handy later.
Now, let's denote by $Q$ the point of intersection of $T_bI$ with perpendicular to $MN$ at $P$ and it remains to prove that $QN$ is indeed parallel to $BI$.
Let the angles $A, B, C$ be $\alpha, \beta, \gamma$.
The angle $BIQ$ is $90^{\circ} - \gamma - \frac{\beta}{2}$

So, in order to prove that $QN || BI$ we need to prove that $\angle T_bQN = 90^{\circ} - \gamma - \frac{\beta}{2}$.
Now, $Q, P, T_b, N$ lie on the same circle with diameter QN, so $\angle T_bQN=\angle T_bPN$ = angle between $BC$ and $PT_b$
Let's do a symmetry about the line $CI$. Under that symmetry the line $BC$ maps to the line $AC$ and the line $PT_b$ maps to $PT_a$.
So, our angle is the angle between $AC$ and $PT_a$. Now, by the "bisector, midline, touchchord " lemma, the line $PT_a$ is just the line $T_cT_a$. 
The angle between $T_cT_a$ and $AC$ is easy to compute. It equals $90^{\circ} - \beta/2 - \gamma$, which finishes the proof.

