Question

Prove that the line joining the midpoints of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half of the difference of the 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)]