If \(E\) is the midpoint of the median \(AD\) of \(\bigtriangleup ABC\), and line \(BE\) intersects \(AC\) at \(F\), show that \(AF\) measures one-third of \(AC\)
Solution:
Given:
In \(△ ABC\), \(AD\) is a median and \(E\) is the mid-point of \(AD\).
Now, Draw \(DP || EF\) as shown in below figure.
In \(△ ADP\), \(E\) is the mid-point of \(AD\) and \(EF || DP\).
Thus, by converse of mid-point theorem, 'The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.'
\(F\) is mid-point of \(AP\).
\(AF=\) --------\((1)\)
In \(△ FBC\),
\(D\) is mid-point of \(BC\) and \(DP || BF\).
So, again by converse of mid-point theorem, 'The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.'
is mid-point of \(FC\).
\(FP=\) ---------\((2)\)
From \((1)\) and \((2)\)
Thus, \(AF = FP = PC\).
\(AC=AF+FP+PC\)
\(AC=3\)
That is, \(AF=\frac{1}{3}AC\)
Hence, proved.