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