Problem 8. In triangle $A B C$, side $A C=42$. The bisector $C L$ is divided by the point of intersection of the bisectors of the triangle in the ratio $2: 1$, counting from the vertex. Find the length of side $A B$, if the radius of the circle inscribed in triangle $A B C$ is 14.
Answer: 56.

Answer: Let $I$ be the center of the inscribed circle in triangle $ABC$ (i.e., the point of intersection of the angle bisectors). Noting that $AI$ is the angle bisector in triangle $ALC$, by the angle bisector theorem, we have: $AC: AL = CI: IL = 2: 1$, from which $AL = AC / 2 = 21$.

Next, $AC \cdot AL \cdot \sin \angle A = 2 S_{\triangle ALC} = 2 S_{\triangle AIC} + 2 S_{\triangle AIL} = AC \cdot r + AL \cdot r = (AC + AL) \cdot r$, where $r$ is the radius of the inscribed circle in triangle $ABC$. Thus, $42 \cdot 21 \cdot \sin \angle A = (42 + 21) \cdot 14$, i.e., $\sin \angle A = 1, \angle A = 90^{\circ}$.

By the angle bisector theorem for $BI$ in triangle $CLB$, we have $BC: BL = CI: IL = 2: 1$. Letting $BL = x$, we have $BC = 2x$. By the Pythagorean theorem: $AC^2 + AB^2 = BC^2$, i.e., $42^2 + (21 + x)^2 = (2x)^2$, from which $x = 35$, and $AB = x + 21 = 56$.