I met an interesting task, tried a lot of ways to prove but didn't find any.
Problem: Let $KLM$ be a triangle, let $I$ be an incircle's center. Let $Q$ be the point on an extension of the segment $KM$ beyond a point $M$; $P$ and $R$ be the points on segments $KL$ and $LM$ respectively such that $P$, $R$ and $Q$ lie on the same line and this line is tangent to incircle of $\triangle KLM$. Circles ($KLM$) and ($KPQ$) intersect again at a point $T$.
Question is how to prove, that $\angle PTI$ is equal to $\angle MTI$?
Notes: Tried to use a fact that ($PLTR$) and ($RTQM$) are also cyclic, prove that perpendiculars to $TP$ and $TM$ from $I$ are equal. So also you can see that incircle tangent points to $RM$ and $KM$, center $I$ and intersection point of perpendicular from $I$ to $TM$ are cyclic with diameter $IM$. Similarly with perpendicular from $I$ to $TP$ and diameter $PI$. Then tried to use trigonometry to prove specified perpendiculars equality. It didn't help.
Let $PK, KM, MR, RP$ be tangent to the incircle at $A, B, C, D$. Writing vectors with origin $I$, the inverses of $K, M, R, P, L, Q$ with respect to the incircle are
$$\vec{K'} = \frac{\vec A + \vec B}{2}, \vec{M'} = \frac{\vec B + \vec C}{2}, \vec{R'} = \frac{\vec C + \vec D}{2}, \vec{P'} = \frac{\vec D + \vec A}{2}, \vec{L'} = \frac{\vec A + \vec C}{2}, \vec{Q'} = \frac{\vec B + \vec D}{2}.$$
Defining
$$\vec X = \frac{\vec A + \vec B + \vec C}{2}, \vec Y = \frac{\vec A + \vec B + \vec D}{2}, \vec Z = \frac{\vec A + \vec B + \vec C + \vec D}{2},$$
we see
$$XK' = XM' = XL' = XZ = YK' = YP' = YQ' = YZ = \frac r2,$$
which is to say that $K'M'L'Z$ lie on a circle and $K'P'Q'Z$ lie on a circle. These circles are the inverses of circles $KML$ and $KPQ$, so $Z$ is the inverse of $T$. Hence $\angle PTI = \angle ZP'I$ and $\angle MTI = \angle ZM'I$, and these are equal because $IM'ZP'$ is a parallelogram.