# 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.

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.

• Could you provide me and other readers a link to the prove of the lemma, that you mentioned at the very beginning? Would be really helpful. – thomas21 Apr 3 at 17:18
• here vk.com/… you can find the proof in Russian (Задача 1), I will translate it to English, but I need some time for that – liaombro Apr 3 at 17:32
• here it is drive.google.com/open?id=1KyH7KDpGsV8qu1x-sBZipxta62QW95dI my English is pretty awful but, I hope, understandable – liaombro Apr 3 at 18:53
• I personally appreciate your efforts very much, +1, thank you. – thomas21 Apr 7 at 17:02