In a trapezium ABCD, if E and F be the mid-point of the diagonals AC and BD respectively. Then, EF = ?
Construction: Join CF and extent it to cut AB at point M
Firstly, in triangle MFB and triangle DFC
DF = FB (As F is the mid-point of DB)
∠DFC = ∠MFB (Vertically opposite angle)
∠DFC = ∠FBM (Alternate interior angle)
∴ By ASA congruence rule
∆MFB ≅ DFC
Now, in triangle CAM
E and F are the mid-points of AC and CM respectively
∴ EF = 1/2 (AM)
EF = 1/2 (AB – MB)
EF = 1/2 (AB-CD)
Hence, option D is correct