Question

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

Answer 1

ANSWER:
In
$△ABC$ we have:
AC = AE + EC ...(i)
AE = EC
... (ii)
∴ AC = 2EC
... (iii)
[ BE is the median of ∆ ABC]
In
$△BEC,DF\parallel BE$
∴ 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)
$\therefore CF=\frac{1}{4}AC$.