In the figure, calculate $QB$ , knowing that $MN=6 $and $NQ=2$ ;moreover, $AB$ , $AN$ and $NB$ are diameters. ($R:QB=4$).
Relations that I found
$\triangle AHN \sim AGB \implies \dfrac{2r}{2R}=\dfrac{NH}{BG} = \dfrac{AH}{2r+GN}$
$\triangle AMB \sim \triangle NHB \implies \dfrac{AM}{NH} = \dfrac{2R}{2r_1} = \dfrac{8+r_1}{BH}$
First of all, it is very easy to realize that $A,N$, and $G$ are co-linear. Let's assume $QB=s$, $\angle NAB=x$,$\angle MAN=y.$ Now, one can easily verify that:
$$\angle QHB= 90^{\circ} -y, \\\angle QBH=90^{\circ} -x-y, \\ \angle QNH=x+y, \\ \angle NHQ=y.$$
So, in $\triangle NHB:$
$$\frac{s}{2}=\frac{\frac{s}{QH}}{\frac{2}{QH}}=\frac {\frac{\cos y}{\cos (x+y)}}{\frac{\sin y}{\sin (x+y)}}=\frac{\cos y \sin (x+y)}{\sin y \cos (x+y)}.$$
On the other hand, in $\triangle MAB$:
$$\frac{\sin (x+y)}{\cos (x+y)}=\frac{8+s}{AM},$$
and in $\triangle MNA$:
$$\frac{\cos y}{\sin y }=\frac{AM}{6}.$$
Thus,
$$\frac{8+s}{6}=\frac{s}{2}\implies s=4.$$
We are done.