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Question

In ΔABC, AD, is a median and E is the mid point of AD. If BE is produced to meet AC in F, show that AF=13AC.

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Solution

AD is median BD=CD

E is mid point of AD

Draw a line DG, such that DG||BF

In ADG
EF||DG and E is mid point of AD

F is midpoint of AG [converse of midpoint theorem]

AF=FG1

In BCF
D is midpoint of BC and DG||BF

FG=GC2

From 1 and 2

AF=FG=GC

AC=AF+FG+GC

13AC=AF


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