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=12(AB+CD)
Proof :- In
ΔGCB,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
∴EF||AB
In
ΔADB,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=12AB
In
ΔBDC,OF||CD and O is the mid-point of BD.
∴ OF=12CD [by converse of mid-point theorem]
On adding Eqs. (i) and (ii), we get
EO+OF=12AB+12CD⇒EF=12(AB+CD)
Hence, proved