Question

In a trapezium ABCD, E and F be the midpoints of the diagonals AC and BD respectively. Then, EF = ?

(a) $\frac{1}{2}AB$

(b) $\frac{1}{2}CD$

(c) $\frac{1}{2}(AB+CD)$

(d) $\frac{1}{2}(AB-CD)$

Answer 1

(d) $\frac{1}{2}(AB-CD)$

Let E and F are midpoints of the diagonals AC and BD of trapezium ABCD respectively.

Draw DE and produce it to meet AB at G.

Consider DAEG and DCED

∠AEG = ∠CED (vertically opposite angles)

AE = EC (E is the midpoint of AC)

∠ECD = ∠EAG (alternate angles)

ΔAEG ≅ ΔCED

⇒ DE = EG → (1)

And AG = CD → (2)

In ΔDGB

E is the midpoint of DG [From (1)]

F is the midpoint of BD

∴ EF is parallel to GB

⇒ EF is parallel to AB

⇒ EF is parallel to AB and CD

Also, EF = $\frac{1}{2}$ GB

⇒EF = $\frac{1}{2}$ (AB − AG)

⇒ EF = $\frac{1}{2}$ (AB − CD) [From (2)]