Question

Question 4
ABC is an isosceles triangle right angled at C. Prove that $A{B}^2=2A{C}^2$.

Answer 1

Given that $\Delta ABC$ is an isosceles triangle right angled at C.
In $\Delta ACB$,
$\angle C={90}^{\circ }$
AC = BC (Given)
$\Rightarrow$ $A{B}^2=A{C}^2+B{C}^2$ ([By using Pythagoras theorem]
$\Rightarrow$ $A{B}^2=A{C}^2+A{C}^2$ [Since, AC = BC]
$\Rightarrow$ $A{B}^2=2A{C}^2$