Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that

The figure is given below:

Given AO =5, BO = 12 and AB = 13

In Δ AOB, AO² + BO² = AB²

∵ 5² + 12² = 13²

25² + 144² = 169²

so by the Pythagoras theorem

Δ AOB is right angled at ∠ AOB.

But ∠ AOB + ∠ AOD forms a linear pair so the given parallelogram is rhombus whose diagonal bisects each other at 90°.