How to show that the distance of the points of tangency along a tangent line on two tangent circles with radius $a$ and $b$ is equal to $2\sqrt{ab}$?

Question

How to show that the distance of the points of tangency along a tangent line on two tangent circles with radius $a$ and $b$ is equal to $2\sqrt{ab}$?

Please see the image below. Line $DE$ is tangent to Circles $B$ and $C$ at point $D$ and $E$, respectively. Line $BC$ passes through point $A$, which is the tangent point of the two given circles. I am trying to prove visually that $DE=2\sqrt{(BA)(AC)}$.

Here is my attempt:

I construct a segment from point $B$ perpendicular to radius $CE$ at point $F$. Since quadrilateral $BFED$ is a parallelogram (a rectangle) $BF=DE$.

Applying the Pythagorean Theorem,

$BF=DE=\sqrt{BC^2-CF^2}$

After this, I got stuck. Any comments or suggestions will be much appreciated.

Answer 1

Draw a paralel to $DE$ through $B$ which cuts $CE$ at $F$. If $DE = x$ then we have, by Pythagora theorem $$x^2+FD^2 = CB^2$$

i.e. $$x^2+(a-b)^2=(a+b)^2\implies x^2=4ab$$

Answer 2

From $B$, draw a parallel line to $DE$. Say, it meets $EC$ at $B'$.

$Sin \theta = \frac{B'C}{AC+AB} = \frac{EC-BD}{AC+AB} = \frac{AC-AB}{AC+AB}$

$DE = (AC+AB)cos\theta = (AC+AB)\sqrt{1-sin^2\theta} = 2\sqrt{AC.AB}$