On the relations between area and bisector of a triangle.

Question

Let us imagine that we have triangle ABC. Now, let us construct a line going from the bisector of the angle B to line segment AC. Now, that line will hit the base precisely at the bisector of the base. Let us name the point of intersection,"D". Therefore, the ratio from AD to DC is 1:1.Also, when we drew the line from point B to line AC, we created two different triangles. What is the relation between the triangles' areas, if there exists one? (Notice: Please answer the question definitively.) Bonus Question: The ratio of AD to DC is currently 1:1. What if we tweaked that angle so that the ratio was, for example, 1:3 or 2:3, how would the area of the resulting triangles change? Can you provide a general formula for the ratio of the triangles' areas for ratio x:y?

Answer 1

Your statement that the ratio from $AD$ to $DC$ is $1:1$ is false. The angle bisector theorem says that if $BD$ is an angle bisector of $\angle ABC$, $\frac{AD}{DC} = \frac{BA}{BC}$, so $AD:DC$ is only $1:1$ if $BA = BC$, which is only true if you have an isosceles triangle. Using trigonometry, a triangle's area is given by $\frac{1}{2} ab \sin C$, so the ratio between the areas of the two triangles is:

$$\frac{1}{2} (BA)(BD) \sin (B/2): \frac{1}{2} (BD)(BC) \sin (B/2)$$ $$= BA:BC$$

You can get the general result using the generalised angle bisector theorem, which is on the same page as the previous link. If $D$ lies on the line $BC$, then:

$$\frac{BD}{DC} = \frac{BA \sin \angle ABD}{BC \sin \angle DBC}$$

Now you can use the formula $A = \frac{1}{2}ab \sin C$ again and find the generalised result.