# What are the possible values ​of the angles of triangle $ABC$?

In a triangle $$ABC$$, let $$AP$$ be the bisector of $$\angle BAC$$ with $$P$$ on the side $$BC$$, and let $$BQ$$ be the bisector of $$\angle ABC$$ with $$Q$$ on the side $$CA$$. We know that $$\angle BAC=60^\circ$$ and that $$AB + BP = AQ + QB$$. What are the possible values ​​of the angles of triangle $$ABC$$?

(Answer: $$\angle ABC = 80^\circ, \angle BCA=40^\circ, \angle BAC=60^\circ$$)

My progress:

The relationships I found:

$$AQ+QB = AB+BP\implies b+x=c+a$$

$$\angle C = 120^\circ -2\theta$$

Considering angle bisector $$BQ$$ of $$\triangle ABC$$: $$\dfrac{c}{b}=\dfrac{AC}{d}$$

Similarly considering $$AP$$: $$\dfrac{c}{a}=\dfrac{b+d}{y}$$

$$\angle 60^\circ +\angle B+\angle C \implies \angle B+\angle C = 120^\circ$$

$$x = \dfrac{(b+d)\cdot c}{AC+c}$$

$$x^2 =\dfrac{(AC)c}{bd}$$

From sine rule:

$$\displaystyle\frac{\sin60}{AC}=\frac{\sin C}{c}=\frac{\sin2\theta}{b+d}=\frac{\sqrt3}{2AC}$$

From cosine rule:

$$c^2 = AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos\angle C$$

$$AC^2 = c^2+BC^2-2c\cdot BC\cdot\cos2\theta$$

...???