### In a triangle ABC, AD is the median. F is the point on AC such that line BF bisect AD at E. Prove that AC = 3AF.

**Answer:**

**Step by Step Explanation:**

- Following picture shows triangle ΔABC with median AD at E, and another line BF, which bisects AD. We have also drawn another line DG which is parallel to BF.

- From proportionality theorem we know that if a line is drawn parallel to one side of a triangle, it divides other two sides in the same ration.
- For triangle ΔBFC, DG is parallel to BF and using proportionality theorem we can find that,

=FG GC BD DC - Since we know that BD = DC, we can find that,

GC = FG ---------(1) - Similarly we can use proportionality theorem for triangle ΔADG,

=AF FG AE ED - But since AE = ED,

FG = AF ---------(2) - Therefore,

AC = AF + FG + GC

Using relation (1) and (2),

AC = AF + AF + AF,

AC = 3AF