Question

In ΔABC, AD is a median and E is the mid-point of AD. If BE is produced to meet AC at F, show that $AF=\frac{1}{3}AC.$

Answer 1

Given: In Δ ABC, AD is a median and E is the mid-point of AD. Also, BE is produced to meet AC at F.
To Prove: $\mathrm{AF}=\frac{1}{3}\mathrm{AC}$
Construction: From D, draw $DG\parallel EF$, meeting AC at G.

Proof:
In ΔBCF, D is the mid-point of BC and $DG\parallel BF$.
∴ G is the mid point of CF.
So, FG = GC
In ΔADG, E is the mid-point of AD and $EF\parallel DG$.
∴ F is the mid-point of AG.
So, AF = FG
Thus, AF = FG = GC
∴ AC = (AF + FG + GC) = 3AF
Hence, $\mathrm{AF}=\frac{1}{3}\mathrm{AC}$