I am not able to use the fact that the smaller circumference is a circumference, what difference does that make to the problem? I've done many, many calculations, but without using this it's impossible to demonstrate the result, how to do it?
Let $R=GO$, $r$ radius of the smallest circle, $x=GH$ and $y$ the distance from O to the point of intersection of the vertical line with the diameter .
Tracing the right triangle inscribed to the semicircle, I used Pythagoras a few times and concluded some results, but that didn't lead anywhere because I don't know how to use that the smaller circle is an inscribed circle, the most promising one is:
$$x^2 = 2R(y+R)$$
Since $P$ is a point of both circles, there is only one tangent vector in $P$, so the points $O, O_2$ and $P$ must be aligned.
As $OO_2 = R-r $, we apply Pythagoras in the triangle $OIO_2$ and conclude that: $$(y+r)^2+r^2 = (R-r)^2 \Rightarrow r^2 +r(2y+2R)+ y^2 -R^2 =0 $$
Solving The equatioin in $r$,we find The positive solution : $$r= -y-R+\sqrt{2R(R+y)}$$
On the other hand, see that:
$$cos(\alpha)=\frac{R+y}{x}=\frac{x}{2R} $$
Then, $x = \sqrt{2R(R+y)} $
Using the first result, we conclude that:
$$r = -y -R +x \Rightarrow x= r+y+R \Rightarrow GH=GI.$$