Question

In an isosceles $\Delta ABC$ right angled at B, prove that $A{C}^2=2A{B}^2.$

Answer 1

In right $\Delta ABC,itisgiventhatAB=BC$ [$\because$ triangle is isosceles] ...(i)

In $\Delta ABC$ by pythagoras theorem,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=A{B}^2+A{B}^2$ [From (i)]
$A{C}^2=2A{B}^2$ Hence Proved.