If  $A=(0,0) B=(2, 1), and C=(3, 0)$are the vertices of a triangle  ABC and BD is  its altitude. The line  through D parallel to the side AB intersects the side BC  at a point K  Then $({\mathrm{\Delta }}_1=\mathrm{Area}\mathrm{of}\mathrm{\Delta ABC},{\mathrm{\Delta }}_2=\mathrm{Area}\mathrm{of}\mathrm{\Delta BDK})$
 

The equation of  side  BC is  $\mathrm{y}-0=\frac{1-0}{2-3}(\mathrm{x}-3)$
 $\mathrm{x}+\mathrm{y}=3....\left(1\right)$
DK is parallel to  AB,  so its equation is
 $\mathrm{y}-0=\frac{1-0}{2-0}(2-2)\Rightarrow \mathrm{x}-2\mathrm{y}=2.....\left(2\right)$
Solving 1and 2  $\mathrm{K}=(\frac{8}{3},\frac{1}{3})$
${\mathrm{\Delta }}_1=\mathrm{Area}\mathrm{of}\mathrm{\Delta ABC}=\frac{1}{2}(\mathrm{AC}\times \mathrm{BD})=\frac{1}{2}\times 3=\frac{3}{2}$
$$\begin{gathered} {\mathrm{\Delta }}_1=\mathrm{Area}\mathrm{of}\mathrm{\Delta BDK}=\frac{1}{2}(\mathrm{BD}\times \mathrm{KL})=\frac{1}{2}\times 1\times \left(\frac{{\displaystyle 8}}{{\displaystyle 3}}-2\right)=\frac{1}{3} \\ \frac{{\mathrm{\Delta }}_1}{{\mathrm{\Delta }}_2}=\frac{9}{2} \end{gathered}$$