The diagonals AC and BD of a parallelogram ABCD intersect at O. If P is the mid-point of AD, prove that

(i) PO || AB

(ii) PO = $\dfrac{1}{2}$CD.

(i) Given,

ABCD is a parallelogram in which diagonals AC and BD intersect each other at O, P is the midpoint of AD.

Join OP.

In parallelogram, diagonals bisect each other,

∴ BO = OD.

Here, O is the mid-point of BD.

In △ABD,

P and O are midpoints of AD and BD respectively,

PO || AB and PO = $\dfrac{1}{2}$AB (By midpoint theorem) ……(i)

Hence, proved that PO || AB.

(ii) ABCD is a parallelogram.

∴ AB = CD …….(ii)

Using both (i) and (ii) we get,

PO = $\dfrac{1}{2}$AB = $\dfrac{1}{2}$CD.

Hence, proved that PO = $\dfrac{1}{2}$CD.