Question

If $\cos \frac{A}{2}=\sqrt{\frac{b+c}{2c}}$, then prove that $a^2+b^2=c^2$

Answer 1

We have,

$\cos \frac{A}{2}=\sqrt{\frac{b+c}{2c}}$

${\cos }^2\frac{A}{2}=\frac{b+c}{2c}$ …….. (1)

Since,

$\cos A=2{\cos }^2\frac{A}{2}-1$

${\cos }^2\frac{A}{2}=\frac{1+\cos A}{2}$

From equation (1)

$\frac{1+\cos A}{2}=\frac{b+c}{2c}$

$1+\cos A=\frac{b+c}{c}$

$\cos A=\frac{b+c}{c}-1$

$\cos A=\frac{b}{c}$

We know that

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

Therefore,

$\frac{b^2+c^2-a^2}{2bc}=\frac{b}{c}$

$\frac{b^2+c^2-a^2}{2b}=b$

$b^2+c^2-a^2=2b^2$

$c^2-a^2=b^2$

$a^2+b^2=c^2$

Hence, proved.