Triangle area division

Question

It is given that in a triangle ABC, a line from A to BC intersects BC at point D. If the ratio in which AD divides BC is given can we say anything about the ratio of areas of triangle ABD and triangle ADC ?

Answer 1

Yes we can find the ratio of the areas of the two triangles.
Area of $\triangle ADC$=$$(AD\times DC\times\sin\theta)\over2$$ Area of $\triangle ABD$=$$(AD\times DB\times \sin(180^\circ-\theta))\over2$$ [$\theta$ is the angle $ADC$]

Divide these two equations and use the ratio you have to get the answer.

Answer 2

It is given that in a $\triangle ABC$, a line from $A$ to $BC$ intersects $BC$ at point $D$. If the ratio in which $AD$ divides $BC$ is given can we say anything about the ratio of areas of $\triangle ABD$ and $\triangle ADC$?

Suppose the area of $\triangle ABD$ is $A_{\triangle ABD}$ and the area of $\triangle ACD$ is $A_{\triangle ACD}$. If the height of the perpendicular line to $BC$ from the point $A$ is $h$:

$$A_{\triangle ABD}=\frac12 h\cdot AD \qquad \qquad A_{\triangle ABD} =\frac12 h\cdot CD$$

$$\therefore \ \frac{A_{\triangle ABD}}{A_{\triangle ABD}} = \frac{\frac12 h\cdot AD}{\frac12 h\cdot CD}=\frac{AD}{ CD} $$