AD and BE are medians of triangle ABC and DF parallel BE.Prove that CF =1fourth AC?

FIRST KNOW THIS PROPERTY - IF THE MID POINTS OF TWO SIDED OF ATRIANGLE ARE JOINT , IT WILL BE PARALLEL TO AND HALF OF THE THIRD SIDE .

E IS THE MID POINT OF AC . SO CE = AC / 2

NOW , D IS THE MID POINT OF BC AND DF IS PARALLEL TO AB . SO , F SHOULD BE MID POINT OF CE .

CF = CE / 2

CE = AC / 2

CF = AC / 4

Given : A ΔPQR in which PS and RT are the medians and SM || RT.

To prove :

Proof : PS and RT are the medians in ΔPQR.

∴ S is mid point of QR and. T is mid point of PQ.

∴ QS = SR and PT = TQ ..(1)

In ΔQTR, S is mid point of QR and SM || RT.

∴ By mid point theorem, M is mid point of QT.

⇒ QM = MT

Thus,

Hi!

In ΔABC, AD and BE are the medians drawn on sides BC and AC respectively.

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

⇒ CE AC and CD BC ...(1)

In ΔEBC,

D is the mid point of BC and DF || BE,

∴ F is the mid point of CE. (Converse of mid point theorem)

⇒ CF EC

⇒ CF × ( AC) (Using (1))

⇒ CF = AC

Cheers!