Question

In the given figure, AD and BE are medians of ΔABC and BE || DF. Prove that CF = 1/4AC.

Answer 1

In ΔBEC,

D is mid-point of BC and from converse of mid-point theorem.

i.e. A line i.e. DF drawn through the mid-point D of BC and parallel to BE bisects the third side i.e. EC at F.

Now, F becomes the mid-point of CE.

⇒ CF = \(\frac { 1 }{ 2 }\) CE

CF = \(\frac { 1 }{ 2 }\) (\(\frac { 1 }{ 2 }\) AC)

[E is mid-point of AC ⇒ AE = EC = \(\frac { 1 }{ 2 }\) AC]

⇒ CF = \(\frac { 1 }{ 4 }\) AC

Hence proved.