Three externally touching circles have their centers on the same line and have radii $a$,$b$ and $c$ (where $a<b<c$).Prove that $b^2=ac$.

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 $b^2=ac$.