Prerequisites:
Area of triangle (proof)
Equidistance property of parallel lines (proof)
Proof:
Let there be a triangle \triangle ABC. Let DE be a line such that DE \parallel BC, where points D and E lie on the sides AB and AC respectively.
In order to prove the theorem, let us join BE and CD by straight lines. Draw EF \bot AB and DG \bot AC.
Now consider the ratio:
\qquad\quad\dfrac{ar (\triangle ADE)}{ar (\triangle BDE)}
Where, ar (\triangle ADE) implies the area of \triangle ADE.
From the figure, using formula for area of triangle, taking EF as a perpendicular on AB,
\qquad\quad\dfrac{ar (\triangle ADE)}{ar (\triangle BDE)} = \dfrac{\frac{1}{2}(AD).(EF)}{\frac{1}{2}(DB).(EF)} = \dfrac{AD}{DB}\qquad\qquad\qquad\qquad\cdots\text{(1)}
Also, by taking DG as a perpendicular on the side AC,
\qquad\quad\dfrac{ar (\triangle ADE)}{ar (\triangle CDE)} = \dfrac{\frac{1}{2}(AE).(DG)}{\frac{1}{2}(EC).(DG)} = \dfrac{AE}{EC}\qquad\qquad\qquad\qquad\cdots\text{(2)}
Again, draw perpendiculars from points D and E on the line BC at the points M and N respectively. Since DE \parallel BC, hence, by equidistance property of parallel lines, DM = EN.
\begin{align}\therefore\quad\;\; ar (\triangle BDC) &= \dfrac{1}{2}(DM)(BC)\\ &= \dfrac{1}{2}(EN)(BC)\\ &= ar (\triangle BEC)\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\cdots\text{(3)}\end{align}
Now, consider area of \triangle BDE,
\qquad\quad\begin{align} ar (\triangle BDE) &= ar (BDEC) - ar (\triangle BEC)\\ &= ar (BDEC) - ar (\triangle BDC)\qquad\qquad\qquad\quad\text{(from $(3)$)}\\ &= ar (\triangle CDE)\qquad\qquad\qquad\qquad\qquad\qquad\quad\cdots\text{(4)}\end{align}
Using the result of (4),
\qquad\quad\dfrac{ar (\triangle ADE)}{ar (\triangle BDE)} = \dfrac{ar (\triangle ADE)}{ar (\triangle CDE)}
Hence from (1) and (2)
\qquad\quad\dfrac{AD}{DB} = \dfrac{AE}{EC}
Hence the result.
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