Question

In $△ABC,\angle A=\frac{\pi }{2}$,prove that $r+2R=\frac{1}{2}(b+c+a)$

Answer 1

$r_1-r=4R{\sin }^2\frac{A}{2}$
.
$r_1=r+4R{\sin }^2\frac{A}{2}$.
Now , $r_1=s.\tan \frac{A}{2}$.
Hence, $s.\tan \frac{A}{2}=r+4R{\sin }^2\frac{A}{2}$
Now it is given that $A=\frac{\pi }{2}$.
Substituting in the above equation, we get
$s.\tan \left(\frac{\pi }{4}\right)=r+4R{\sin }^2\left(\frac{\pi }{4}\right)$
$s=r+2R$
$\frac{a+b+c}{2}=r+2R$.