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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?

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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, $|CD|=|EC|=4$

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

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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, $$|DC|=4$$

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