Saudi Arabian Mathematical Competitions

On the side $BC$ take point $M$ such that $CM = AB'$. The angle bisector theorem implies
$$\frac{AB'}{B'C} = \frac{AB}{BC} = \frac{AC}{BC} \text{, hence } \frac{MC}{B'C} = \frac{AC}{BC} \text{.}$$

It follows $\frac{MC}{AC} = \frac{B'C}{BC}$, that is $\triangle MCB' \sim \triangle ACB$, and we get $MC = MB'$. Moreover, $\widehat{C} = \widehat{MCB'} = \widehat{MB'C}$ and $MC = MB'$.
From $BB' + B'A = BC$ it follows $BB' = BC - B'A = BC - MC = BM$, hence $\triangle B'BM$ is isosceles.
In $\triangle BB'M$ we have $180^{\circ} = 2\widehat{C} + 2\widehat{C} + \frac{\widehat{C}}{2} = \frac{9\widehat{C}}{2}$, implying $\widehat{C} = 40^{\circ}$. It follows
$$\widehat{B} = \widehat{C} = 40^{\circ} \text{ and } \widehat{A} = 100^{\circ} .$$

In $\triangle ABB'$, we have
$$\frac{BB'}{\sin 4x} = \frac{AB'}{\sin x} = \frac{AB}{\sin 3x} .$$
In $\triangle BB'C$ we have
$$\frac{BB'}{\sin 2x} = \frac{B'C}{\sin x} = \frac{BC}{\sin 3x}$$
and in $\triangle ABC$ we can write
$$\frac{BC}{\sin 4x} = \frac{AC}{\sin 2x} .$$

We get
$$BB' + BB' \frac{\sin x}{\sin 4x} = BB' \frac{\sin 3x}{\sin 2x}$$
hence
$$\begin{equation*} 1 + \frac{\sin x}{\sin 4x} = \frac{\sin 3x}{\sin 2x} \tag{1} \end{equation*}$$
Relation (1) is equivalent to
$$\frac{\sin x}{\sin 4x} = \frac{\sin 3x - \sin 2x}{\sin 2x},$$
that is
$$\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin 2x \cos 2x} = \frac{2 \sin \frac{x}{2} \cos \frac{5x}{2}}{\sin 2x} .$$
We get
$$\cos \frac{x}{2} = 2 \cos 2x \cos \frac{5x}{2},$$
or
$$\cos \frac{x}{2} = \cos \frac{9x}{2} + \cos \frac{x}{2} .$$
It follows $\cos \frac{9x}{2} = 0$, that is $\frac{9x}{2} = \frac{\pi}{2}$, and we obtain $x = \frac{\pi}{9}$.
Finally,
$$\widehat{B} = \widehat{C} = \frac{2\pi}{9}, \quad \widehat{A} = \frac{5\pi}{9} .$$