Show that $\frac{1}{\sqrt{r_3}}=\frac{1}{\sqrt{r_1}}+\frac{1}{\sqrt{r_2}}$ for three mutually tangent circles, each tangent to a common line 
Three circles are tangents to the line $AB$. Being $r_1$ the radius of the biggest one, $r_2$ the radius of the middle one and $r_3$ the radius of the smallest. Show that $$\dfrac{1}{\sqrt{r_3}}=\dfrac{1}{\sqrt{r_1}}+\dfrac{1}{\sqrt{r_2}}.$$

Hint:show that $AB = 2\sqrt{r_1r_2}$.

I know I have to use the Pitagora`s Theorem. 
 A: Let $A$ and $B$ be tangency points to the biggest circle and to the middle circle respectively. 
Also, let $O_1$, $O_2$ and $O_3$ be centers of circles with radius $r_1$, $r_2$ and $r_3$ respectively 
and let $O_3K$ and $O_3M$ be perpendiculars from $O_3$ to $O_1A$ and $O_2B$ respectively.
Thus, $O_1K=r_1-r_3$, $O_1O_3=r_1+r_3$ and by the Pythagoras's theorem we obtain:
$$O_3K=\sqrt{(r_1+r_3)^2-(r_1+r_3)^2}=2\sqrt{r_1r_3}.$$
Similarly $$O_3M=2\sqrt{r_2r_3}$$ and
$$AB=2\sqrt{r_1r_2}$$ and since $AB=O_3K+O_3M$, we obtain:
$$2\sqrt{r_1r_2}=2\sqrt{r_1r_3}+2\sqrt{r_2r_3}$$ or
$$\frac{1}{\sqrt{r_3}}=\frac{1}{\sqrt{r_1}}+\frac{1}{\sqrt{r_2}}.$$
Done!
A: Let $C$ be the tangency point of the middle circle. If you could show that $AB = 2\sqrt{r_1r_3},$ then similarly you can show that $AC = 2\sqrt{r_1r_2}$ and $CB = 2\sqrt{r_2r_3}$ and use the fact that $AB = AC+CB.$ Your result will follow immediately. 
As for how to prove the hint, look at the trapezoid $ABSL$, where $S$ and $L$ are the centers of the smaller and the larger circles respectively. 
