In the given figure. ABCD is a trapezium in which AB|| DC and E is the mid-point of AD, if EF||DC, then show that `EF=(1)/(2)(AB+DC)`.

Explanation

Join AC which interests EF at G.
In `triangleADC, " "EG"||"DC`
and E is the mid-point of AD
then G is mid-point of AC `" "` (converse of mid-point theorem)
Again, E is mid point of AD.
Since, G is mid-point of AC,
`EG =(1)/(2) DC" "` (mid point theorem)...(1)
Now, EF"||" AB
`" "` EF"||"AB
In `triangleCAD, `
Since, G is mid-point of CA,
and `" "GF"||"AB" "(becuase EF"||"AB)`
`therefore ` F is mid point of CB. `" "`(converse of mid-point theorem)
In `triangleCAB`,
Since, F is mid point of CB.
and G is mid-point of CA.
`therefore GF=(1)/(2)AB" "`(mid-point theorem)...(2)
Adding (1) and (2), we get
`EG+GF=(1)/(2)(CD+AB)`
`implies " "EF=(1)/(2)(AB+CD)` Hence proved.