Question

In the given figure, $△$ ABC and $△$ DBC are on the same base BC. If AD and BC intersect each other,

Prove that $\frac{Area\left(△ABC\right)}{Area\left(△DBC\right)}=\frac{AO}{DO}$

Answer 1

$\Rightarrow$ Draw $AL$ perpendicular to $BC$ from point $A$

and $DM$ perpendicular to $BC$ from point $D$.

In $△ALO$ and $△DMO$

$\Rightarrow$ $\angle AOL=\angle DMO$ [ Vertically opposite angles ]

$\Rightarrow$ $\angle ALO=\angle DMO$ [ Both are right angles ]

$\therefore$ $△ALO\sim △DMO$ [ By AA criteria ]

$\frac{ar\left(△ALO\right)}{ar\left(△DMO\right)}=\frac{(AO{)}^2}{(DO{)}^2}=\frac{(AL{)}^2}{(DM{)}^2}$ [ Ratio of the area of two similar triangle is equal to the squares of the corresponding sides ]

$\therefore$ $\frac{(AO{)}^2}{(DO{)}^2}=\frac{(AL{)}^2}{(DM{)}^2}$ or

$\frac{AO}{DO}=\frac{AL}{DM}$ ----- ( 1 )

Now, Area of $△ABC=\frac{1}{2}\times BC\times AL$

Area of $△DBC=\frac{1}{2}\times BC\times DM$

$\Rightarrow$ $\frac{ar\left(△ABC\right)}{ar\left(△DBC\right)}=\frac{\frac{1}{2}\times BC\times AL}{\frac{1}{2}\times BC\times DM}=\frac{AL}{DM}=\frac{AO}{DO}$ [ From ( 1 ) ]

$\therefore$ $\frac{ar\left(△ABC\right)}{ar\left(△DBC\right)}=\frac{AO}{DO}$