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In a triangle ABC, AD is a median and E is the midpoint of median AD. A line through B and E meets AC at point F. Prove that $AC = 3AF$

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Hint:For this question mid-point theorem is to be used.
Mid-point theorem: - It says a line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the length of the third side.
Example: -
                          
In triangle ABC, D is midpoint of AB and E is midpoint of AC.
$
\therefore DF//BC\\
\therefore DF = \dfrac{1}{2}BC $
So we can say that it bisects the line into two equal parts.

Complete step-by-step answer:
Construction: - Draw DG parallel to BF, which meets FC at G.
                  
Proof: - In triangle ABC, AD is median and E is the midpoint of AD.
In and E is the midpoint of AD. Hence, F is midpoint of AG (midpoint theorem)
Or, $AF = FG$ ................equation(1)
In and D is the midpoint of BC.
$\therefore $ G is midpoint of CF (midpoint theorem)
Or, $FG = GC$ .................equation(2)
From equation (1) and (2)
$AF = FG = GC$
Since: - $AC = AF + FG + GC$
$AC = 3AF$
Hence proved.

Note: Carefully observe the hint which is given in the question and then proceed for the question. Here $DG\parallel BF$ gives a hint that the mid- point theorem is to be used. Similar questions can be framed with the use of a converse mid-point theorem.
Converse mid-point theorem: - It states that when a line is drawn through the midpoint of a side of a triangle which is parallel to the second side then it will bisect the third side.
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