Question

In a rhombus ABCD, prove that : $A{C}^2+B{D}^2=4B{C}^2$

Diagonals of a rhombus bisect each other at right angle.

Answer 1

Given: Rhombus ABCD. Diagonals meet at O
In a rhombus, diagonals biects each other at a right angle.
hence, $OA=OC=\frac{1}{2}AC$
$OB=OD=\frac{1}{2}BD$
In $△OBC$,
$O{B}^2+O{C}^2=B{C}^2$
$(\frac{1}{2}BD{)}^2+(\frac{1}{2}AC{)}^2=B{C}^2$
$B{D}^2+A{C}^2=4B{C}^2$