Question

In a Rhombus ABCD the diagonals bisect at O prove that AC square + BD square equal 4AB square

Answer 1

Refer to the attached image.

In a rhombus, all the sides are of equal measure and diagonals of the rhombus bisect each other at right angle.

Consider the rhombus ABCD,

where AB = BC = AD = CD.

Consider right angled triangle AOB,

By Pythagoras theorem, we get

$(Hypotenuse)^2 = (base)^2 + (height)^2$

$(AB)^2 = (OB)^2 + (OA)^2$

$(AB)^2 = (\frac{1}{2}BD)^2 + (\frac{1}{2}AC)^2$

$(AB)^2 = (\frac{BD^2}{4}) + (\frac{AC^2}{4})$

$4AB^2 = {BD^2} +{AC^2}$

Hence, proved.