Answer :


Given: In ΔABC, the AD is a median and E is mid-point of the AD and BE is produced to meet AC in F.

To Prove:

Construction: Draw DG || BF as shown in figure


Now, In ΔBFC

DG || BF [By construction]

As AD is a median on BC, D is a mid-point of BC


G is a mid-point of CF [By mid-point theorem]

CG = FG …[1]

Now, In ΔADG

EF || DG [By Construction]

As E is a mid-point of AD [Given]


F is a mid-point of AG [By mid-point theorem]

FG = AF …[2]

From [1] and [2]

AF = CG = FG …[3]


AC = AF + FG + CG

AC = AF + AF + AF [From 3]

AC = 3AF

Hence Proved

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