The correct option is
A True
Given:
△ABC, E is mid point of median AD.
Construction: Draw
DG∥BFIn
△ADG,
EF∥DG and E is mid point of AD
Hence, F is mid point of AG (Mid point theorem)
or,
AF=FG...(I)
In
△BCF,
DG∥BF and D is mid point of BC
Therefore, G is mid point of CF (Mid point theorem)
or,
FG=GC...(II)
From I and II
AF=FG=GCSince,
AC=AF+FG+GCHence,
AC=3AF