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

Open in App

Solution

In $△ A B C$ ,

By Pythagoras Theorem,

$(A B)^{2} = (A C)^{2} + (B C)^{2} . . . . . (1)$

Since, $△ A B C$ is an isosceles triangle,

$∴$ $A C = B C . . . . (2)$

$∴$ From $(1)$ and $(2),$

$(A B)^{2} = 2 (A C)^{2}$ $[h e n c e p r o v e d]$

Suggest Corrections

Similar questions

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

A B^{2} = 2 A C^{2}

{A B}^{2} = 2 {A C}^{2}

A B^{2} = 2 A C^{2}

△ A B C

Join BYJU'S Learning Program