Question

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

Answer 1

In $△$AOB,
AO = 5 cm
OB = 12 cm
AB = 13 cm
5, 12 and 13 form a Pythagorean triplet.
Thus, $△$AOB is a right angle triangle, right angled at O.
ABCD is a parallelogram.
So, diagonals bisect each other.
$\Rightarrow$AO = OC = 5 cm
Also, OB = OD = 12 cm.
Thus, in parallelogram ABCD, diagonals bisect at right angles.
Hence, ABCD is a rhombus.