Question

In the given figure AD and BE are the medians of ∆ABC 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 =

1

2

CE

CF =

1

2

(

1

2

AC)

[E is mid-point of AC ⇒ AE = EC =

1

2

AC]

⇒ CF =

1

4

AC

Hence proved.