Proof that inscribed circle's radius is always $\sqrt{{ab}}$

Question

In image $AD$ is diameter of circle , $MB$ and $NC$ are perpendicular to $AD$ , $AB=a$ and $CD=b$. And small circle touches these perpendiculars, diameter and big circle

I have found that radius of small circle is always equal to $\sqrt{{ab}}$ , but I can't find an exact way to prove it.How would you prove it?

Answer 1

Denote the tangency points by $X,Y,Z$ according to the following picture.

First prove that $Y$ lies on $AX$ and that $Z$ lies on $DX$.

Then observe that $\triangle ABY \sim \triangle ZCD$. Conclude that $\dfrac{AB}{BY} = \dfrac{CZ}{CD}$. This means that $ab=AB \cdot CD = BY \cdot CZ$. However, it is clear that $BY=CZ=r$ where $r$ is the radius of the small circle, so $r=\sqrt{ab}$.