ABCD is a trapezium and P and Q are the mid-points of the diagonals of AC and BD. Then PQ is equal to :

1. $\dfrac{1}{2}$ AB

2. $\dfrac{1}{2}$ CD

3. $\dfrac{1}{2}$ (AB − CD)

4. $\dfrac{1}{2}$ (AB + CD)

The Mid-point Theorem, which states that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of its length.

In triangle ADC,

M is the mid-point of AD (let)

P is the mid-point of AC

MP ∥ CD

MP = $\dfrac{1}{2}$ CD

In triangle ABD,

M is the mid-point of AD

Q is the mid-point of BD

MQ ∥ AB

MQ = $\dfrac{1}{2}$ AB

Since: MP ∥ CD, MQ ∥ AB and AB ∥ CD,

∴ MP ∥ MQ

Both pass through point M. So MP and MQ are the same straight line.

Hence, M, P and Q all lie on same line.

MQ = MP + PQ

PQ = MQ - MP

PQ = $\dfrac{1}{2}$ AB - $\dfrac{1}{2}$ CD

PQ = $\dfrac{1}{2}$ [AB - CD]

Hence, option 3 is the correct option.