Question

Prove that, "If the angles of a triangle are ${45}^o-{45}^o-{90}^o$, then each of the perpendicular sides is $\frac{1}{\sqrt{2}}$ times the hypotenuse."

Answer 1

Given that the angles of the triangle are ${45}^0-{45}^0-{90}^0$

Let the corresponding sides be $a,b,c$

By sine rule, we have $\frac{a}{\sin \left({45}^0\right)}=\frac{b}{\sin \left({45}^0\right)}=\frac{c}{\sin \left({90}^0\right)}$

$\Rightarrow \sqrt{2}a=\sqrt{2}b=c$

$\Rightarrow a=b=\frac{c}{\sqrt{2}}$

Therefore, each of perpendicular sides is $\frac{1}{\sqrt{2}}$ times the hypotenuse.

Hence, proved.