Let $a,b,c$ be the radii of the circles $(A),(B),(C),$ respectively. Assuming that $(C)$ is the smallest circle, then we have the relation $\sqrt{ab}=\sqrt{ac}+\sqrt{ab}$ $(\star)$ (well-known). Label $\angle ACB=\gamma.$ Then by cosine law in $\triangle ABC,$ we have $\cos \gamma=\frac{c(a+b+c)-ab}{(a+c)(b+c)} \Longrightarrow \gamma > 90^{\circ} \Longleftrightarrow \ c(a+b+c)-ab<0$ Thus, using the relation $(\star),$ the latter inequality is equivalent to $c\left [ c-2\sqrt{ab} + \left (\sqrt{a}+\sqrt{b} \right )^2 \right ] <c \left ( \sqrt{a}+\sqrt{b} \right )^2 \Longleftrightarrow c < 2\sqrt{ab}$ Which is clearly true, since the length of the common tangent between $(A),(B)$ is greater that the radius of $(C).$ Now, using $\cos \gamma$ from the latter expression, we get $\sin \frac{\gamma}{2}= \sqrt{\frac{1-\cos \gamma}{2} }=\sqrt{\frac{ab}{(a+c)(b+c)}}= \sqrt{\frac{1}{ \left ( 1+\frac{c}{a} \right) \left (1+ \frac{c}{b} \right)}}$ So $\sin \frac{{\gamma}}{2}$ is maximum $\Longleftrightarrow$ $\Delta=\left ( 1+\frac{c}{a} \right) \left (1+ \frac{c}{b} \right)$ is minimum. Setting $x=\sqrt{ \frac{_c}{^a}},$ $y=\sqrt{ \frac{_c}{^b}},$ we shall minimize $\Delta=(1+x^2)(1+y^2)$ with the constraint $x+y=1,$ coming from the equation $(\star).$ Therefore, letting $\omega=xy,$ we obtain $\Delta=(1+x^2)(1+y^2)=\omega^2-2\omega+2$ where $\frac{_1}{^4} \ge \omega >0,$ due to $\omega=xy=x(1-x)=x-x^2$ with $1>x>0.$ The quadratric $\omega^2-2\omega+2$ is strictly decreasing in the interval $\frac{_1}{^4} \ge \omega >0,$ so the minimum value of $\Delta$ occurs when $\omega=\frac{_1}{^4}$ $\Longrightarrow$ $x=y=\frac{_1}{^2}$ $\Longrightarrow$ $\sin \frac{_{\gamma}}{^2}$ reaches its maximum value when $a=b,$ i.e. $(A) \cong (B).$ $\sin \frac{\gamma}{2}=\sqrt{\frac{a^2}{(a+c)^2}}= \frac{a}{ a+\frac{a}{4}} =\frac{4}{5} \Longrightarrow \ 90^{\circ} < \gamma \le 2 \sin^{-1} \left (\frac{_4}{^5} \right) \approx 106,26^{\circ}.$