Question

In the adjoining figure, AD and BE are the medians of ∆ABC and DF || BE. Show that $CF=\frac{1}{4}AC$.

Answer 1

In ∆ABC, we have:
AC = AE + EC ...(i)
AE = EC ...(ii) [BE is the median of ∆ABC]
∴ AC = 2EC ...(iii)

​In ∆BEC, DF || BE.
∴ EF = CF (By midpoint theorem, as D is the midpoint of BC)
But EC = EF + CF
⇒ EC =​ 2 ⨯ ​CF ...(iv)
From (iii) and (iv), we get:
AC = 2 ⨯ (2 ⨯​ CF)
∴​ CF = $\frac{1}{4}AC$