Question

The base BC of triangle ABC is divided at D so that BD = $\frac{1}{2}$ DC.
Prove that area of $\Delta ABD=\frac{1}{3}$ of the area of $\Delta ABC$.

Answer 1

Let $BD=x$ then $BC=3x$ Given BD=$\frac{1}{2}$DC

We know area of triangle = $\frac{1}{2}base\times hight$

In triangle ABC

Area of triangle ABC=$\frac{1}{2}BC\times hight$ =$\frac{1}{2}3x\times h$ Let height of triangle $ABC=h$

And area of triangle ABD=$\frac{1}{2}x\times h$ Let height of triangle $ABD=h$. Because height is common in both triangle.

Then area of triangle ADB=$\frac{xh}{2}$

And area of triangle ABC=$\frac{3xh}{2}$

Then $Areaof\Delta ABD=\frac{1}{3}Areaof\Delta ABC$