Question

Prove that the sum of the squares on the sides of a rhombus is equal to the sum of square on its diagonals.

Answer 1

Given: ABCD is a rhombus with side length AB = BC = CD = DA = a. Diagonals AC and BD meet at O.

To Prove: Sum of the squares on the sides equals the sum of the squares on the diagonals.

i.e. AB² + BC² + CD² + DA² = AC² + BD²

or equivalently 4a² = AC² + BD²

Proof (Step-wise):

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

So, `AO = CO = (AC)/2` and `BO = DO = (BD)/2`.

2. Consider right triangle AOB since AO ⟂ BO.

By the Pythagorean theorem

AB² = AO² + BO²

3. Multiply that equation by 4:

4·AB² = 4(AO² + BO²) 

= (2AO)² + (2BO)²

= AC² + BD²

4. Since all four sides of the rhombus are equal.

AB² + BC² + CD² + DA² = 4·AB²

Combining with step 3 gives

AB² + BC² + CD² + DA² = AC² + BD²

Hence, the sum of the squares on the sides of a rhombus equals the sum of the squares on its diagonals.