In parallelogram ABCD, E is the mid-point of AB and AP is parallel to EC which meets DC at point O and BC produced at P. Prove that :

(i) BP = 2AD

(ii) O is mid-point of AP.

By mid-point theorem,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

By converse of mid-point theorem,

The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side.

(i) In △ ABP,

⇒ E is the mid-point of AB and EC || AP.

∴ C is the mid-point of BP. (By converse of mid-point theorem)

∴ BP = 2BC ………(1)

Since, ABCD is a parallelogram.

∴ AD = BC (Opposite sides of parallelogram are equal) …….(2)

From equation (1) and (2), we get :

⇒ BP = 2AD.

Hence, proved that BP = 2AD.

(ii) Since, opposite sides of parallelogram are parallel.

∴ AB || CD

⇒ AB || OC.

In △ ABP,

⇒ E is the mid-point of AB and OC || AB.

∴ O is the mid-point of AP. (By converse of mid-point theorem)

Hence, O is the mid-point of AP.