In a trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.

Prove that :

AB + DC = 2EF.

Join BE and produce to meet CD produced at point P.

In △ PDE and △ BAE,

⇒ ∠PED = ∠BEF (Vertically opposite angles are equal)

⇒ AE = ED (Since, E is the mid-point of AD)

⇒ ∠EDP = ∠EAB (Alternate angles are equal)

∴ △ PDE ≅ △ BAE (By A.S.A. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BE = EP and AB = PD.

In △ BPC,

Since, E and F are mid-points of sides BP and BC respectively.

∴ EF = $\dfrac{1}{2}PC$.

To prove :

AB + CD = 2EF ……..(1)

Substituting value in L.H.S. of equation (1), we get :

⇒ AB + CD = PD + CD = PC.

Substituting value in R.H.S. of equation (2), we get :

⇒ 2EF = $2 \times \dfrac{1}{2}PC$ = PC.

Since, L.H.S. = R.H.S.

Hence, proved that AB + CD = 2EF.