Let ABCD be a trapezium in which AB||DC and let E be the midpoint of AD. Let F be a point on BC such that EF||AB. Prove that

(i) F is the midpoint of BC, (ii) `EF =(1)/(2)(AB+DC).`

Explanation

Join BD, cutting EF at M.
Now, in `triangle`DAB, E is the midpoint of AD and EM||AB.
`therefore ` M is the midpoint of BD.
`therefore EM=(1)/(2) AB " " `…(i)
Again, in `triangle` BDC, M is the midpoint of BD and MF||DC.
`therefore` F is the midpoint of BC.
`therefore MF=(1)/(2) DC. " " `...(ii)
`therefore EF= EM+MF =(1)/(2) (AB+DC) " " `[from (i) and (ii) ].
Hence, F is the midpoint of BC and `EF =(1)/(2) (AB+DC).`