ABC is a triangle. AD is a median of ABC. E is the midpoint of AD and produced to meet AC at F. prove that AF=1/3AC

**Given:** AD is the median of ΔABC. E is the mid point of AD. BE produced meets AD at F

**To Prove :**

**Construction: **From Point D, draw DG BF.

**Proof: **In ΔADG, E is the mid-point of AD and EF||DG.

∴F is the mid point of AG [Converse of the mid point theorem]

⇒ AF = FG ... (i)

In ΔBCF, D is the mid point of BC and DG||BF

∴ G is the mid point of CF

⇒ FG = GC ... (ii)

From (i) and (ii), we get,

AF = FG = GC ... (iii)

Now, AF + FG + GC = AC

⇒ AF + AF + AF = AC [Using (iii)]

⇒ 3AF = AC

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