Question

AD and BD are medians of triangle ABC and BE is parallel to DF. Prove that CF = AC/4

Answer 1

please clarify the question. I meant please send the real question.this dosent look like a valid question.

Answer 2

Step-by-step explanation:

Given that in a triangle ABC, AD and BE are medians of a triangle ABC.  

So D and E are the mid points of the sides BC and AC respectively.

So BD = DC  and AE = EC ........1

Now in triangle ΔEBC,   since D is the mid point of BC and DF || BE,

then by mid point theorem,

F is the mid point of CE.  i.e. CF = EF and EC = 2CF ........2  

Now AC = AE + EC  => AC = AE + EF + CF  => AC = EC + CF + CF  => AC = 2CF + CF +CF  => AC = 4CF  => CF = AC/4

So corect answer is CF = AC/4

Hope it helps you...

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