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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)]
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)]
Khumanshintry:
I don't know
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