Question

In $△ABC,AD$ is the mediam through $A$ and $E$ is the midpoint of $AD.BE$ produced meets $AC$ in $F$ such that $BF\parallel DK$, prove that $AF=\frac{1}{3}AC$

Answer 1

Given AD is the median of $\Delta ABC$ and E is the midpoint AD

Through D, draw DS||EF in $\Delta ADS$, E is the midpoint of AD and EF || DS By converse of midpoint theorem we have F is midpoint of AS & AF = FS _______ (1)

Similarly as $\Delta BCF$

D is the midpoint of BC and DS||BF

S is the midpoint of CF and FS=SC __________ (2)

AF=FS=SC __________ (3)

AF+FS+SC=AC

3AF=AC $\Rightarrow AF=\frac{1}{3}AC$