Question

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = `1/2` (AB + CD).

[Hint: Join BE and produce it to meet CD produced at G.]

Answer 1

Given: ABCD is a trapezium in which AB || CD. Also, E and F are respectively the mid-points of sides AD and BC.

Construction: Join BE and produce it to meet CD produced at G, also draw BD which intersects EF at O.

To prove: EF || AB and EF = `1/2` (AB + CD).

Proof: In ΔGCB, E and F are respectively the mid-points of BG and BC, then by mid-point theorem,

EF || GC

But GC || AB or CD || AB   ...[Given]

∴ EF || AB

In ΔADB, AB || EO and E is the mid-point of AD.

Therefore by converse of mid-point theorem, O is mid-point of BD.

Also, EO = `1/2` AB  ...(i)

In ΔBDC, OF || CD and O is the mid-point of BD.

∴ OF = `1/2` CD   [By converse of mid-point theorem] ...(ii)

On adding equations (i) and (ii), we get

EO + OF = `1/2` AB + `1/2` CD

⇒ EF = `1/2` (AB + CD)

Hence proved.