## Maximum Size of Square Inscribed in Isosceles Triangle

The diagram shows a square of side
$x$
inscribed in an isosceles triangle of height
$h$
and base angle
$\alpha$
.

From the diagram
$tan \alpha = \frac{x}{a}\rightarrow a= \frac{x}{\tan \alpha}$

Also
$tan \alpha = \frac{h}{a+ x/2} = \frac{h}{x/tan \alpha + x/2}= \frac{2h tan \alpha}{2x+ xtan \alpha}$

From the last equation,
$1=\frac{2h}{2x+ xtan \alpha} \rightarrow x(2+ tan \alpha) =2h \rightarrow x= \frac{2h}{2+ tan \alpha}$