Consider an isosceles triangle

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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 $d =\sqrt {r(r-2p)}$.

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

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This is another masterpiece of Euler. This is general result of what you have asked.

Source: H.S.M. Coxeter and S.L. Greitzer- Geometry Revisited.

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Consider the following figure:

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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.