Answer :

Proof:


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



Proof:


Now, In ΔBFC


DG || BF [By construction]


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


Therefore,


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]


Therefore,


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


FG = AF …[2]


From [1] and [2]


AF = CG = FG …[3]


And


AC = AF + FG + CG


AC = AF + AF + AF [From 3]


AC = 3AF



Hence Proved


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