Consider an isosceles triangle 
Consider an isosceles triangle. Let $r$ be the radius of its circumscribed circle and $p$ the radius of its inscribed circle. Prove that the distance $d$ between the centres of these two circles is $d =\sqrt {r(r-2p)}$.


I could not get any idea to solve. However I have tried to make a figure (partially).

Consider the following figure:

Using Euclid's theorem of sides in a right triangle one has
$$2r(r+d+p)=b^2=4r^2-a^2=4r^2-\left({2r\over r+d}\>p\right)^2\ .$$
It follows that
$$(r+d)^2(r+d+p)=2r\bigl((r+d)^2-p^2\bigr)=2r(r+d+p)(r+d-p)\ .$$
Removing the factor $r+d+p$ leads to
$$r^2+2rd+d^2=2r(r+d-p)\ ,$$
from which the claim immediately follows.
