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Question
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.
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Solution
Given: In a ∆ABC, AD is a median and E is the mid-point of AD.
Construction: Draw DP || EF.
Proof: In ∆ADP, E is the mid-point of AD and EF || DP.
So, F is mid-point of AP. ...[By converse of mid-point theorem]
In ∆FBC, D is mid-point of BC and DP || BF.
So, P is mid-point of FC
Thus, AF = FP = PC
∴ AF = `1/3` AC
Hence proved.
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