Prove that the circle which contains ATB and incircle of ABC touch in one point(T). Incircle of $ABC$ touches $AC$ in $D$, $BC$ in $E$ and $AB$ in $K$. $J$ is the center of the excircle which touches the side $AB$. The circumcircle of $ADJ$ and $BEJ$ intersect in point $J$ and $T$. Prove that the circumcircle of $ATB$ and incircle of $ABC$ touch in one point.
I got that $T$ has to be on incircle of $ABC$. It also looks like that $K$ is on $JT$. One thing which could help is that $T$, center of the incircle of $ABC$ and center of the circumcircle of $ATB$ (labeled $M$ below) are colinear.


First, observe that $\angle DTE = \angle DTJ + \angle JTE = \left(\pi - \angle JAD\right) + \left(\pi - \angle EBJ\right) = 2\pi - \angle JAD - \angle EBJ = 2\pi - \left(\frac \pi 2 + \frac \alpha 2\right) - \left(\frac \pi 2 + \frac \beta 2\right) = \pi - \frac \alpha 2 - \frac \beta 2.$
On the other hand, $\angle EKD = \pi - \angle DKA - \angle BKE = \pi - \left(\frac \pi 2 - \frac \alpha 2\right) - \left(\frac \pi 2 - \frac \beta 2\right) = \frac \alpha 2 + \frac \beta 2$.
Thus $\angle DTE + \angle EKD = \pi$ which implies the fact you mentioned: $T$ lies on the incircle $o$ of triangle $ABC$.
Note that $\angle ATJ = \angle ADJ = \pi - \angle JDC = \pi - \angle CEJ = \angle JEB = \angle JTB$, so $TJ$ is the angle bisector of angle $ATB$.
Notice that $DK \perp AI \perp AJ$, so $DK \parallel AJ$. Moreover, $\angle TJA = \pi - \angle ADT = \angle TDC = \angle TKD$. Therefore $TK \parallel TJ$, so $K \in TJ$. Since $TJ$ is the angle bisector of angle $ATB$, we deduce that $\angle ATK = \angle KTB$.
Let $TK$ intersect circumcircle $\omega$ of triangle $ATB$ at points $T$ and $X$. Then $K$ is the midpoint of arc $AB$ of $\omega$ because $TK$ is bisector of angle $ATB$. Let $l$ be the tangent line to $\omega$ at $X$. Then $AB \parallel l$. 
Consider homothety centered at $T$ which maps $X$ to $K$. Then $l$ is mapped to $AB$ so $\omega$ is mapped to a circle passing through $T$ which is tangent to line $AB$ at point $K$ - this circle is $o$. As a consequence, $\omega$ and $o$ are tangent to each other at point $T$. 
