Question

ans this que plzz with full method

Answer 1

given
AD is the median of ABC and E is the midpoint of AD
through D
draw DG II BF
In ADG
E is the midpoint of AD and EF II DG
by converse of midpoint Theorem we have F is a midpoint of AG and AF =FG....... (1)
similarly in BCF
D is the midpoint of BC and DG II BF
G is the midpoint of CF and FG = GC...... (2)
from equations (1) and (2)
we will get
AF = FG = GC ...... (3)

AF+FG+GC=AC
AF+AF+AF=AC
FROM eu (3)
3AF = AC
AF = (1/3) AC