In a trapezium ABCD, AB // DC, E is mid-point of AD and F is mid-point of BC, then :

1. 2EF = $\dfrac{1}{2}(AB + DC)$

2. 2EF = AB + DC

3. EF = AB + DC

4. EF = $\dfrac{1}{2} \times AB \times DC$

Join AC. Let AC intersects EF at point O.

We know that,

In trapezium the line joining the mid-points of non-parallel sides are parallel to the parallel sides of trapezium.

∴ AB || EF || DC.

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,

⇒ EF || DC

⇒ EO || DC

In △ ADC,

E is mid-point of AD and EO || DC.

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

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

Given,

⇒ EF || AB

⇒ OF || AB

In △ ABC,

O is mid-point of AC and F is mid-point of BC.

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

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

⇒ EO + OF = $\dfrac{1}{2}DC + \dfrac{1}{2}AB$

⇒ EF = $\dfrac{1}{2}(DC + AB)$

⇒ 2EF = AB + DC.

Hence, Option 2 is the correct option.