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,  
\[1=\frac{2h}{2x+ xtan \alpha} \rightarrow x(2+ tan \alpha) =2h \rightarrow x= \frac{2h}{2+ tan \alpha}\]
 

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