Question

In a trapezium ABCD, if E and F be the mid-point of the diagonals AC and BD respectively. Then, EF = ?
(a) $\frac{1}{2}AB$
(b) $\frac{1}{2}CD$
(c) $\frac{1}{2}(AB+CD)$
(d) $\frac{1}{2}(AB-CD)$

Answer 1

(d) $\frac{1}{2}\left(AB-CD\right)$

Explanation:

Join CF and produce it to cut AB at G.
Then ∆CDF ≅ ∆GBF [∵ DF = BF, ∠​DCF = ∠​BGF and ∠​CDF = ∠​GBF]
∴ CD = GB
Thus, in ∆​CAG, the points E and F are the mid points of AC and CG, respectively.
∴ EF = $\frac{1}{2}\left(AG\right)$ = $\frac{1}{2}\left(AB-GB\right)=\frac{1}{2}\left(AB-CD\right)$