How prove $G,H,T $ are collinear. Question:
Circle $O_{1}$ and $O_{2}$ are internally tangent at point $T$. $AB$ and $CD$  are tangents of circle $O_{1}$, the angle bisectors of Angle $\angle ADB$  and $\angle CBD$ intersects at point $G$, while the angle bisectors of angle $\angle CDB$ and $\angle ABD$ intersect at point $H$. 
Prove that :
$G,H,T $ are collinear. 

Thank you for you help,
This is china TST Training supplementary questions.2014
 A: Here's an outline of a very tedious coordinate-based approach.

Things can start out rather promisingly. Center the big circle, say of radius $r$ at the origin, and place $T$ at $(r,0)$. Abusing notation to write "$\operatorname{cis}\theta$" for "$(\cos\theta, \sin\theta)$", assign these coordinates:
$$A = \operatorname{cis}4\alpha \qquad B = \operatorname{cis}4\beta \qquad C = \operatorname{cis}4\gamma \qquad D = \operatorname{cis}4\delta$$
with $0 < \beta < \alpha < \pi/2$ and $0 < \gamma < \delta < \pi/2$.
If the radius of the small circle is $s$, so that $O_2 = (r-s,0)$, then the tangency condition requires that 
$$\operatorname{dist}(O_2,\overline{AB})=s=\operatorname{dist}(O_2,\overline{BC}) \qquad\qquad (\star)$$
Extend $\overline{BG}$ and $\overline{BH}$ to meet the circle again at $B^\prime$ and $B^{\prime\prime}$; extend $\overline{DG}$ and $\overline{DH}$ to meet the circle again at $D^\prime$ and $D^{\prime\prime}$. Conveniently, these new points bisect arcs determined by $A$, $B$, $C$, $D$, and we have
$$B^\prime = \operatorname{cis}2(\gamma+\delta)\qquad B^{\prime\prime} = \operatorname{cis}2(\alpha+\delta) \qquad D^\prime = \operatorname{cis}2(\alpha+\beta) \qquad D^{\prime\prime} = \operatorname{cis}2(\gamma+\beta)$$
Then $G = \overline{BB^\prime} \cap \overline{DD^\prime}$, while $H = \overline{BB^{\prime\prime}} \cap \overline{DD^{\prime\prime}}$, and collinearity with $T$ is equivalent to this condition
$$\frac{ G_x - T_x }{G_y - T_y} = \frac{ H_x - T_x }{ H_y - T_y} \qquad \text{or, more simply,} \qquad H_y ( G_x - r ) = G_y ( H_x - r ) \qquad (\star\star)$$
One then checks that the collinearity condition $(\star\star)$ is compatible with the tangency condition $(\star)$. Easy, right?
Unfortunately, while the fundamental points ---$A$, $B$, $C$, $D$, and $B^\prime$, $B^{\prime\prime}$, $D^\prime$, $D^{\prime\prime}$--- are uncomplicated, equations $(\star)$ and especially $(\star\star)$ are less-so, taking some trigonometric massaging to arrive at the key relation that seems to make everything work:
$$\tan\alpha \tan\gamma = \tan\beta \tan\delta \qquad\quad (\star\star\star)$$
Even with a computer algebra system such as Mathematica, this is an ordeal.

I suspsect that a keen insight leads to an elegant synthetic proof. The OP's diagram includes points $I_1$ and $I_2$, which (along with $G$ and $H$) complete the set of incenters of the four possible triangles determined by $A$, $B$, $C$, $D$; that they're shown to be collinear with points of tangency $E$ and $F$ (a fact that's not obvious to me) is no-doubt a hint about how to proceed, but I haven't seen a way to make use of that hint. (Maybe it'll come to me immediately after I post this answer. Edit: Nope.)
