TITLE: If $J$ is tangent point of $GH$ with incircle of $FGH$ and $D$ is intersection of $F$-mixtilinear inclrcle with $(FGH)$, then $\angle FGH=\angle GDJ$.
QUESTION [11 upvotes]: Let $FGH$ be a triangle with circumcircle $A$ and incircle $B$, the latter with touchpoint $J$ in side $GH$. Let $C$ be a circle tangent to sides $FG$ and $FH$ and to $A$, and let $D$ be the point where $C$ and $A$ touch, as shown here.

Prove that $\angle FGH = \angle GDJ$.

REPLY [4 votes]: Using the Evan Chen's mixtilinear incircle article in here, the results become trivial.
I will change some notations.
In $\triangle ABC$, let the incircle hit $BC$ at $D$ and the $A$-mixtilinear incircle hit the circumcircle of $\triangle ABC$ at $E$. Prove that $\angle DEB = \angle ABC$. 
Since (9) holds, we have $\angle DTM_A = \angle AFB=180-\angle B - \frac{1}{2}\angle A$, where $F = AI \cap BC$.
Since $\angle BTM_A = 180-\frac{1}{2} \angle A$, we have $$\angle DTB = \angle BTM_A - \angle DTM_A = \angle B = \angle ABC$$ as desired.