Question

In given figure $△$ABC and $△$BDC are on the same base BC

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

Answer 1

In the given figure, construct $AM$ and $DN$ such that $△ABC$ and $△BDC$ are on the same base $BC$ with $AM\perp BC$ and $DN\perp BC$ as shown in the above figure:

We know that the area of a triangle is $A=\frac{1}{2}\times b\times h$ where $b$ is the base and $h$ is the height of the triangle.

Here, area of $△ABC$ is $A_1=\frac{1}{2}\times BC\times AM$ and area of $△BDC$ is $A_2=\frac{1}{2}\times BC\times DN$

Dividing both the areas, we get

$\frac{A_1}{A_2}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times BC\times DN}=\frac{AM}{DN}.......\left(1\right)$

Now, consider $△AOM$ and $△DON$,

$\angle AMO=\angle DNO={90}^0$ (Construction)

$\angle AOM=\angle DON$ (Vertically opposite angles)

Therefore, $△AOM\sim △DON$ (A similarity criterion)

Thus,

$\frac{AM}{DN}=\frac{AO}{DO}.......\left(2\right)$

Substitute equation $\left(2\right)$ in equation $\left(1\right)$ as follows:

$\frac{A_1}{A_2}=\frac{AM}{DN}=\frac{AO}{DO}$

$\Rightarrow \frac{Area(△ABC)}{Area(△BDC)}=\frac{AO}{DO}$ $\left[henceproved\right]$