Answer:

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

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

In △CDF and △GBF

⇒ DF=BF [ Since, F is mid-point of diagonal BD ]

⇒ ∠DCF=∠BGF [ DC∥GB and CG is a transversal ]

⇒ ∠CDF=∠GBF [ DC∥GB and BD is a transversal ]

∴ △CDF≅△GBF [ By ASA congruence rule ]

⇒ CD=GB [ C.P.C.T ] ---- ( 1 )

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

∴ EF=

2

1

(AG)

⇒ EF=

2

1

(AB−GB)

From ( 1 )

⇒ EF=

2

1

(AB−CD)