Three externally touching circles have their centers on the same line and have radii $a$,$b$ and $c$ (where $aThree circles have their centers on the same line and have radii $a$,$b$ and $c$ (where $a<b<c$).The circle with radius $b$ touches the other two circles but circles with radii $a$ and $c$ do not touch each other.The three circles also have a common tangent.Prove that $b^2=ac$.

I solved it with good deal of calculations. Is there an elementary way to do this problem
 A: We are given a circle $A$ with radius $a$ externally tangent to a circle
$B$ with radius $b>a$. The two common tangents $T_1,T_2$ of $A$ and $B$ meet in point $P$. On the other side of $B$ from $A$ is the circle $C$ with radius $c>b$ externally tangent to $B$.
Let $T$ be a similarity (or homothetic) map with center at $P$ which maps
the center of $A$ to the center of $B$. The two common tangents $T_1,T_2$
are fixed by $T$ since they each contain $P$. Since $A$ and $B$ are externally tangent, $T$ must map $B$ to a circle externally tangent to $B$ and with tangents $T_1,T_2$. Thus that circle must be $C$.
Because ratios are unchanged by similarities, the ratios $b/a$ and $c/b$
must be equal, hence $b^2=ac$.
A: 
Let $AK=x$. By similar triangles $\triangle AHE \sim \triangle AID \sim \triangle AJB$:
$$\frac {x+a}a=\frac {x+2a+b}b=\frac{x+2a+2b+c}c$$
Subtracting $1$ from each fraction:
$$\frac xa = \frac{x+2a}b=\frac {x+2a+2b}c=:y$$
We have $$\frac{ay+2a}b=y = \frac{by+2b}c$$
$$acy+2ac = b^2y+2b^2$$
Factoring out $y+2$ gives the result.
A: Say common tangent length between circle A and B (IJ) is $x$, B and C (JK) is $y$ and so between A and C (IK) is $(x + y)$.

Applying Pythagoras, between circles of radius a and b,
$x^2 = (a+b)^2 - (b-a)^2 = 4ab$
Applying Pythagoras, between circles of radius b and c,
$y^2 = (b+c)^2 - (c-b)^2 = 4bc$
Applying Pythagoras, between circles of radius a and c,
$(x+y)^2 = (a+2b+c)^2 - (c-a)^2 = 4(b^2+ab+bc+ca)$
i.e $x^2 + y^2 + 2xy = 4b^2+4ca+x^2+y^2$
i.e $4b^2+4ca = 2xy = 8b\sqrt{ac}$
i.e $b^2 - 2b\sqrt{ac} + ca = 0$
i.e $b^2 = ca$
A: Elementary solution :
The top two right triangles in your diagram are similar. So
$$ \frac{height}{base} = \frac{b-a}{b+a} = \frac{c-b}{c+b}  $$
By componendo and dividendo,
\begin{align}
 \frac{b}{a} & = \frac{c}{b} \\
 b^2 & = ac
\end{align}
This aside, @Somos' answer is great. Since all circles are similar, one can get any circle by dilating or contracting any other circle. The ratio of dilation is just the ratio of radii (radius is the only parameter of a circle).
In your diagram, add other direct common tangent to these three circles. The two tangents meet at P, which clearly is center of similarity (homothety).

Fun fact :
Whenever there are tangent circles inscribed in an angle like given, the radii of circles form geometric progression.
When the number of circles is any odd number, the circle exactly in the middle has radius equal to the geometric mean of radii of two circles at the end.
Do you see why this is obvious? :)
