Question

in a rhombus ABCD the diagonals bisect at O prove that AC square + BD square equal to 4 a b square

Answer 1

Hence it is proved ........

Answer 2

Answer:

The proof is explained below :

Step-by-step explanation:

To prove : AC² + BD² = 4AB²

Proof : ABCD is a rhombus

AB = BC = CD = AD

$OC=\frac{1}{2}AC..............(1)\\\\OB=\frac{1}{2}BD.............(2)$

∠BOC = 90° (Diagonals of rhombus bisect each other at 90°)

In ΔBOC , By Pythagoras thorem,

BC² = OB² + OC²

$BC^2=(\frac{1}{2}\cdot BD)^2+(\frac{1}{2}\cdot AC)^2\text{...........from (1) and (2)}\\\\BC^2=\frac{BD^2+AC^2}{4}\\\\\implies4\cdot BC^2=BD^2+AC^2..........(3)\\\text{Now, BC=AB=CD=AD So, from equation (3)}\\4\cdot AB^2=AC^2+BD^2$

Hence Proved