3. The angle $A C B$ of triangle $A B C$ has a size of $140^{\circ}$. The bisector of angle $A B C$ intersects side $A C$ at point $X$. Point $Y$ lies on side $A B$ such that angle $Y C B$ has a size of $100^{\circ}$. Determine the size of angle $Y X B$.
Solution. Let $P$ denote the orthogonal projection of point $X$ onto line $BC$ (Fig. 3). Since $|\varangle XCY| = 40^{\circ} = |\varangle XCP|$, point $X$ lies not only on the bisector of angle $ABC$ but also on the bisector of angle $YCP$. Therefore, point $X$ has the same distance from the three lines $AB$, $BC$, and $CY$, so it also lies on the bisector of angle $AYC$, i.e., $|\varangle AYX| = |\varangle AYC| / 2$. If we denote $\beta = |\varangle ABC|$, then $|\varangle CYB| = 80^{\circ} - \beta$, $|\varangle AYC| = 100^{\circ} + \beta$, and $|\varangle AYX| = 50^{\circ} + \frac{\beta}{2}$, thus $|\varangle XYB| = 130^{\circ} - \frac{\beta}{2}$. From $\triangle XYB$, it finally follows that $|\varangle YXB| = 180^{\circ} - \left(130^{\circ} - \frac{\beta}{2}\right) - \frac{\beta}{2} = 50^{\circ}$.