Question

In the figure, $ ABC$ and $ DBC$ are two triangles on the same base $ BC$. If $ AD$ intersects $ BC$ at $ O$, show that $ \frac{area (△ABC) }{area (△DBC)}=\frac{AO}{DO}$

Answer 1

Given :

1. $BC$ is common base to triangles $ABC$ and $DBC$.
2. $AD$ intersects $BC$ at $O$.

Construction: Draw $AP\perp BC$ and $DM\perp BC$

To prove : $\frac{area(△ABC)}{area(△DBC)}=\frac{AO}{DO}$

Proof:

Area of a triangle $=\frac{1}{2}\times$ base $\times$ height

$\Rightarrow$$area\left(△ABC\right)=\frac{1}{2}\times BC\times AP$$...\left(i\right)$

$\Rightarrow$$area\left(△DBC\right)=\frac{1}{2}\times BC\times DM$$...\left(ii\right)$

Consider $△APO$ and $△DMO$

$\angle APO=\angle DMO$ …[each is equal to $90°$]

$\angle AOP=\angle DOM$ …[vertically opposite angles]

Hence by AA similarity criteria

$△APO$$~$$△DMO$

In similar triangles the ratio of corresponding sides is equal

$\Rightarrow$$\frac{AP}{DM}=\frac{AO}{DO}$$...\left(iii\right)$

Consider the ratio of area of triangles $ABC$ and $DBC$

$\frac{area(△ABC)}{area(△DBC)}=\frac{{\displaystyle \frac{1}{2}}\times BC\times AP}{{\displaystyle \frac{1}{2}}\times BC\times DM}$

$\Rightarrow$$\frac{area(△ABC)}{area(△DBC)}=\frac{AP}{DM}$

From $\left(iii\right)$

$\Rightarrow$$\frac{area(△ABC)}{area(△DBC)}=\frac{AO}{DO}$

Hence proved.