Question

In Fig. AD and BE are medians of ΔABC and BE || DF. Prove that CF $=\frac{1}{4}AC.$

Answer 1

In ΔBEC, DF is a line through the mid-point D of BC and parallel to BE intersecting CE at F. Therefore, by converse of mid-point theorem F is the mid-point of CE.

Now, F is the mid-point of CE

$\Rightarrow CF=\frac{1}{2}CE$

$\Rightarrow CF=\frac{1}{2}\left(\frac{1}{2}AC\right)[EismidpointofACi.e.EC=\frac{1}{2}AC]$

$\Rightarrow CF=\frac{1}{4}AC$

Hence proved