Question

The diagonals of a rhombus ABCD intersect each other at O. Prove that: `OA^2 + OC^2 = 2AB^2 - 1/2 BD^2`.

Answer 1

Given: ABCD is a rhombus; diagonals AC and BD intersect at O.

To Prove: `OA^2 + OC^2 = 2AB^2 - 1/2 BD^2`

Proof [Step-wise]:

1. In a rhombus, the diagonals bisect each other at right angles.

So, AO = OC and BO = OD.

Denote AO = OC = x and BO = OD = y.

2. Consider side AD.

In right triangle AOD.

By Pythagoras AD²

= AO² + OD²

= x² + y²

But all sides of the rhombus are equal.

So, AD = AB and hence AB² = x² + y².

3. Compute the left-hand side:

LHS = AO² + OC²

= x² + x²

= 2x²

4. Compute the right-hand side:

BD = 2·BO = 2y

So, BD² = 4y². 

`RHS = 2 AB^2 - 1/2 BD^2` 

= `2(x^2 + y^2) - 1/2 (4y^2)` 

= 2x² + 2y² − 2y²

= 2x²

5. LHS = RHS both equal 2x². 

So, the required identity holds.

Therefore, `OA^2 + OC^2 = 2AB^2 - 1/2 BD^2`, as required.