The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that :

PR = $\dfrac{1}{2}$ (AB + CD)

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.

Given,

⇒ PR // AB

⇒ PQ // AB

In △ ABD,

P is mid-point of AD and PQ // AB.

∴ Q is mid-point of BD. (By converse of mid-point theorem)

∴ PQ = $\dfrac{1}{2}AB$ (By mid-point theorem) ……….(1)

Given,

⇒ PR // DC

⇒ QR // DC

In △ BCD,

Q is mid-point of BD and QR // DC.

∴ R is mid-point of BC. (By converse of mid-point theorem)

∴ QR = $\dfrac{1}{2}CD$ (By mid-point theorem) ……….(2)

Adding equations (1) and (2), we get :

⇒ PQ + QR = $\dfrac{1}{2}AB + \dfrac{1}{2}CD$

⇒ PR = $\dfrac{1}{2}(AB + CD)$.

Hence, proved that PR = $\dfrac{1}{2}(AB + CD)$.