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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