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 $2sqrt{ab}$?

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 $2sqrt{ab}$?
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 $DE=2sqrt{(BA)(AC)}$.

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.

Math Lover · Answer

From $B$, draw a parallel line to $DE$. Say, it meets $EC$ at $B'$.
$Sin theta = frac{B'C}{AC+AB} = frac{EC-BD}{AC+AB} = frac{AC-AB}{AC+AB}$
$DE = (AC+AB)costheta = (AC+AB)sqrt{1-sin^2theta} = 2sqrt{AC.AB}$

Aqua · Answer

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

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 $2sqrt{ab}$?

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