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

The correct option is D $\frac{1}{2}(AB-CD)$

$ABCD$ is a trapezium and $E,F$ are mid-points of diagonal $AC$ and $BD$

$AB\parallel CD$ [ one par of opposite side is parallel in trapezium ]

In $△CDF$ and $△GBF$

$\Rightarrow$ $DF=BF$ [ Since, $F$ is mid-point of diagonal $BD$ ]

$\Rightarrow$ $\angle DCF=\angle BGF$ [ $DC\parallel GB$ and $CG$ is a transversal ]

$\Rightarrow$ $\angle CDF=\angle GBF$ [ $DC\parallel GB$ and $BD$ is a transversal ]

$\therefore$ $△CDF\cong △GBF$ [ By $ASA$ congruence rule ]

$\Rightarrow$ $CD=GB$ [ C.P.C.T ] ---- ( 1 )

In $△CAG,$ the points $E$ and $F$ are the mid-points of $AC$ and $CG$ respectively.

$\therefore$ $EF=\frac{1}{2}\left(AG\right)$

$\Rightarrow$ $EF=\frac{1}{2}(AB-GB)$

From ( 1 )

$\Rightarrow$ $EF=\frac{1}{2}(AB-CD)$