AD and BE are medians of triangle ABC and DF parallel BE.Prove that CF =1fourth AC?
Let us observe the following figure.
Consider the triangle ABC.
AD and BE are the medians of the triangle.
Thus, D is the midpoint of the side BC and E is the midpoint of the side AC.
Consider the triangle BEC.
Given that DF is parallel to BE.
Therefore, F is the midpoint of CE.
Thus, it has been proved that
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
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)
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