Question

5. In the figure, D is the midpoint of side
BC of ∆ABC. G is the midpoint of seg
AD. seg BG when produced meets
side AC at F. Prove AF =1/3 AC using
basic proportionality theorem.
[Hint : Draw line parallel to BF passing through D]​

Answer 1

Answer:

bf is parallel to de (given) and d is mid point of bc (given) so by converse of mid point theorem e is the mid point of ac

FE=EC. ....... ......1

similarly AF=FE... ..............2

from equations 1&2

AF=EC=FE

AF+EC+FE = AC

3AF=AC (as we proved that AF=EC=FE)

AF=1/3AC