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  • E is the mid-point of a median AD of ABC and BE is produced to meet AC at F. Show that AF=1/3 AC

E is the mid-point of a median AD of \triangle ABC and BE is produced to meet AC at F. Show that AF=\frac{1}{3}AC .

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Solution.
Given: In \triangle ABC, AD is a median and E is the mid-point of AD
To Prove: AF=\frac{1}{3}AC.
Construction: Draw DP\parallel EF and \triangle ABC as given

Proof: In \triangle ADP, E is mid-point of AD and EF\parallel DB
So, F is mid-point of AP {converse of midpoint theorem}
In \triangle FBC, D is mid-point of BC and DP\parallel BF
So, P is mid-point of FC {converse of midpoint theorem}
Thus, AF = FP = PC
AF+FP+PC=AC=3AF
\therefore AF=\frac{1}{3}AC
Hence Proved

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