Question

In the figure,$△ABC$ and $△DBC$ are on the same base $BC$. $AD$ and $BC$ intersect at $O$. Prove that $\frac{area(∆ABC)}{area(∆DBC)}=\frac{AO}{DO}$.

Answer 1

Step 1: Check the similarity of $△AOB$ and $△COD$

Given that, $△ABC$ and $△DBC$ are on the same base $BC$. $AD$ and $BC$ intersect at $O$.

AAA similarity: If in two triangles, the corresponding angles are equal, then the triangles are similar.

In $△AOB$ and $△COD$,

$\angle AOB=\angle DOC$ [Vertically opposite angle]

$\angle ABO=\angle CDO$ [Altarnate angles]

$\angle OAB=\angle OCD$ [Altarnate angles]

According to AAA similarity

$△AOB~△COD$

Similar Triangles property: If two triangles are similar, then their corresponding sides are proportional.

Since $△AOB~△COD$, so

$\frac{AB}{DC}=\frac{OA}{OD}=\frac{OB}{OC}$ ….. (1)

Step 2: Finding the ratio of areas of $△ABC$ and $△DBC$

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

The height of $△ABC$ is $AB$.

The height of $△DBC$ is $DC$.

Area of $△ABC=\frac{1}{2}\times base\times height=\frac{1}{2}\times BC\times AB$…(2)

Area of $△DBC=\frac{1}{2}\times base\times height=\frac{1}{2}\times BC\times DC$…(3)

By dividing equations (2) and (3), we get

$\frac{\text{Area of}△ABC}{\text{Area of}△DBC}=\frac{{\displaystyle \frac{1}{2}}\times BC\times AB}{{\displaystyle \frac{1}{2}}\times BC\times DC}$

$\Rightarrow \frac{\text{Area of}△ABC}{\text{Area of}△DBC}=\frac{AB}{DC}$

From (1) we get

$\frac{\text{Area of}△ABC}{\text{Area of}△DBC}=\frac{OA}{OD}$

Hence it is proved that $\frac{area(∆ABC)}{area(∆DBC)}=\frac{AO}{DO}$.