In Figure, `A B C D` isa trapezium in which side `A B` is a parallel to side `D C` and `E` is the mid-point of side `A Ddot` If `F` is a point on the side `B C` such that the segment `E F` is parallel to side `D C` . Prove that `F` is the mid point of `B C` and `E F=1/2(A B+D C)dot`

Explanation

Given ABCD is a trapezium in which AB||CD. Also, E and F are respectively the mid-points of sides AD and BC.

Construction Join BE and produce it to meet CD produced at G, also draw BD which intersects EF at O.
To prove EF||AB and EF `=(1)/(2)`(AB+CD).
Proof In `Delta`GCB, E and F are respectively the mid-points of BG and BC, then by mid-point theorem,
`" "EF||GC`
But `" "GC||AB or CD||AB" "[given]`
`therefore" "EF||AB`
In `Delta`ADB, AB||EO and E is the mid-point of AD.
Therefore by converse of mid-point theorem, O is mid-point of BD.
Also, `" "EO=(1)/(2)AB" "...(i)`
In `Delta`BDC, OF||CD and O is the mid-point of BD.
`therefore" "OF=(1)/(2)CD" "` [ by converse of mid-point theorem]...(ii)
On addillng Eqs. (i) and (ii), we get
`" "EO+OF=(1)/(2)AB+(1)/(2)CD`
`rArr" "EF=(1)/(2)(AB+CD)" "` Hence proved.