Answer :

The figure is given below:

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

In Δ AOB, AO^{2} + BO^{2} = AB^{2}

∵ 5^{2} + 12^{2} = 13^{2}

25^{2} + 144^{2} = 169^{2}

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°.**

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