Question

Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel
to each of the parallel sides and is equal to half of the difference of these sides
​

Answer 1

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 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 = ½ GB

⇒EF = ½ (AB − AG) ⇒ EF = ½ (AB − CD) [From (2)]