Question

AD and BE are medians of ∆ABC. DF||BE. Find AC:CF​

Answer 1

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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 /2AC)

[E is mid-point of AC ⇒ AE = EC = 1/2 AC]

⇒ CF = 1/4 AC

Hence proved.

hope its help u