## Radius of Circle Inscribed in Isosceles Triangle

A circle of maximum size inscribed in an isosceles triangle will touch the sides of the triangle tangentially at three points as shown in the diagram below.

From the diagra AC is common to triangles ABC and ACD, both tirnagles are right angled and have corresponding sides equal - BC and CD both equal the radius of the inscribed circle. Both triangles are congruent therefore, and have equal angles
$\alpha$
.
Then
$tan \alpha = \frac{r}{x}$
and
$tan 2 \alpha = \frac{h}{x}$

Use the identity
$tan 2 \alpha = \frac{2 tan \alpha}{1- tan^2 \alpha}$
to obtain
$\frac{h}{x} = \frac{2r/x}{1- r^2/x^2}$
.
This equation can be rearranged to give
$r^2 h + 2rx^2-hx^=0$

The solutions to this equation are
$r=-x \pm \sqrt{x^2+h^2}$

Obviously only the positive option is possible, so
$r=-x + \sqrt{x^2+h^2}$