question_answer

In a trapezium ABCD, if E and F be the mid-points of the diagonals, AC and BD, respectively. Then, EF is equal to

A) \[\frac{1}{2}AB\]                      

B) \[\frac{1}{2}CD\]

C) \[\frac{1}{2}(AB+CD)\]             

D) \[\frac{1}{2}(AB-CD)\]

Correct Answer: D

Solution :

Join CF and produce it to meet AB at G.

Then, \[\Delta CDF\cong \Delta GBF\]                   [by AA criterion]

CD = GB and CF = GF

Since, E and F are mid-points of CA and CG, we have

\[EF=\frac{1}{2}AG=\frac{1}{2}\,\,(AB\,-GB)=\frac{1}{2}\,\,(AB-CD)\]