Question

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

Answer 1

Given $BD=DC=\frac{1}{2}BC$

$AE=ED=\frac{1}{2}AD$

$BX$ is parallel to $BF$.

Consider traiangle $BFC$ and $DKC$

$\angle BFC=\angle DkC$, $\angle FCD=\angle KCD$

$BFC$ and $DKC$ are similar .(AAA similarity)

& $\frac{FC}{KC}=\frac{BC}{DC}=\frac{BC}{\frac{1}{2}BC}=2$

$\Rightarrow \frac{FC}{KC}=2$ $\Rightarrow KC=\frac{1}{2}FC$

$\text{K is mid point of FC}$ $FK=KC$ ....(1)

Consider triangles $AEF$ and $AKD$

$\angle EAF=\angle DAK$ $\angle BFA=\angle DKA$ (size $BF||DK$)

$\therefore \Delta AFE$ & $\Delta AKD$ are similar (AAA similirity )

$\Rightarrow \frac{AF}{AK}=\frac{AE}{AD}=\frac{AE}{2AE}=\frac{1}{2}$

$\Rightarrow AF=\frac{AK}{2}$

$\Rightarrow F$ is mid point of $AK$

$\Rightarrow AF=FK$ ...(2)

From 1 and 2

$FK=KC=AF$

$\therefore \frac{AF}{AC}=\frac{AF}{AF+FK+KC}=\frac{AF}{3AP}=\frac{1}{3}$

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