# Angle Bisectors of triangles.

I was working on a geometry problem relating to the angle bisectors of triangles :

In triangle $$\Delta ~ABC$$, $$~∠A=40°,~ ∠B=20°,~$$ and $$~AB − BC = 4~$$. Find the length of angle bisector from $$~∠C~$$.

I was able to figure out a majority of the angle measures, but I was unable to utilize the information about the side lengths to find the angle bisector.

Does anyone what method I have to use to solve this?

Let the angle bisector intersects with $$|AB|$$ on $$D$$ and take a point, $$E$$ on $$|AB|$$ such that $$\angle{ECB}=80$$ $$\,$$ Then, $$\,$$ $$|BE|=|BC|$$,$$\,$$ $$\,$$ $$|AE|=4$$ $$\,$$ $$\angle{ECA}=40$$ $$\,$$ which gives us $$|EC|=4$$. Also $$\angle{ECD}=20$$, $$\angle{DEC}=80$$ $$\,$$ and$$\,$$ $$\angle{EDC}=80$$ $$\,$$ Thus, $$|CD|=|EC|=4$$

Second solution by using sine rule: Let the angle bisector intersects with |AB| on D

$$\dfrac{sin{C}}{sin{A}}=\dfrac{sin{120^{\circ}}}{sin{40^{\circ}}}=\dfrac{sin({40^{\circ}}+{80^{\circ}})}{sin{40^{\circ}}}=cos80^{\circ}+2cos^240^{\circ}=1+2cos80^{\circ}=\dfrac{x+4}{x}=1+\dfrac{4}{x}$$

$$\\$$ Thus, $$\dfrac{4}{x}=2cos80$$

$$\\$$

Again by using sine rule on $$\triangle{BCD}$$ , $$\dfrac{x}{|DC|}=\dfrac{sin{80^{\circ}}}{sin{20^{\circ}}}=\dfrac{sin{80^{\circ}}}{sin{160^{\circ}}}=\dfrac{1}{2cos{80^{\circ}}}$$

$$\\$$ Thus, $$|DC|=4$$