In the Figure 5.43, ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively.

Then prove that, PQ || AB and

Given AB ∥ DC

P and Q are mid points of AD and BC respectively.

Construction :- Join AC

The figure is given below:

In Δ ADC

P is mid point of AD and PQ is ∥ DC the part of PQ which is PO is also ∥ DC

By mid=point theorem

A line from the mid-point of a side of Δ parallel to third side, meets the other side in the mid-point

⇒ O is mid-point of AC

⇒ PO = 1/2 DC…………..1

Similarly in Δ ACB

Q id mid-point of BC and O is mid –point of AC

⇒ OQ∥ AB and OQ = 1/2 AB………………2

Adding 1 and 2

PO + OQ = 1/2 (DC+ AB)

PQ = 1/2 (AB +DC)

And PQ ∥ AB

Hence proved.