11.1. In triangle $\mathrm{ABC}$, the bisectors of angles $\mathrm{BAC}$ and $\mathrm{BCA}$ intersect sides ВС and АВ at points К and Р, respectively. It is known that the length of side АС is equal to the sum of the lengths of segments АР and СК. Find the measure of angle $\mathrm{ABC}$.
Answer: $60^{\circ}$.

Solution 1. Let the angles of triangle ABC be denoted by the corresponding letters $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$, and the intersection point of the angle bisectors by I. Reflect points P and K with respect to the angle bisectors AK and CP, respectively; their images will be points $\mathrm{P}^{\prime}$ and $\mathrm{K}^{\prime}$ on side $\mathrm{AC}$ such that $\mathrm{AP}^{\prime}=\mathrm{AP}, \mathrm{CK}^{\prime}=C K$. Since $\mathrm{AP}^{\prime}+\mathrm{CK}^{\prime}=\mathrm{AP}+\mathrm{CK}=\mathrm{AC}$, points $\mathrm{P}^{\prime}$ and $\mathrm{K}^{\prime}$ coincide, and we denote this point as M. The measures of angles AMI=API=APC $=180^{\circ}-\mathrm{A}-$ C/2, CMI=CKI=CKA $=180^{\circ}-\mathrm{C}-\mathrm{A} / 2$. Their sum is $360^{\circ}-3(\mathrm{~A}+\mathrm{C}) / 2=360^{\circ}-$ $270^{\circ}+3 \mathrm{~B} / 2=90^{\circ}+3 \mathrm{~B} / 2=180^{\circ}$, from which $\mathrm{B}=60^{\circ}$. Therefore, the measure of angle $\mathrm{ABC}$ is 60.

Solution 2. Let the lengths of sides $\mathrm{AB}, \mathrm{BC}, \mathrm{AC}$ of the triangle be $c, a, b$ respectively, and find the lengths of all segments used in the problem using the property of the angle bisector. Then $A P=\frac{b c}{a+b}, C K=\frac{a b}{b+c}$, from which $A P+C K=\frac{b c}{a+b}+\frac{a b}{b+c}=\frac{b^{2} c+b c^{2}+a^{2} b+a b^{2}}{a b+a c+b c+b^{2}}=A C=b$. Simplify both sides by $b$, multiply by the denominator, combine like terms, and we get: $b^{2}=a^{2}+c^{2}-a c$. By the cosine rule, $b^{2}=a^{2}+c^{2}-2 a c \cdot \operatorname{Cos} B$, from which $\operatorname{Cos} B=\frac{1}{2}$ and $B=\angle A B C=60^{\circ}$.

Grading criteria. In the first solution ($\bullet$) The idea of reflecting points P and K with respect to the angle bisectors AK and CP is present: 1 point, ($\cdot$) It is shown that the images under such reflection will give a common point on side AC: another 1 point. In the second solution ($\cdot$) AP and CK are found: 3 points. ($\cdot$) The equality $b^{2}=a^{2}+c^{2}-a c$ is obtained: another 2 points. ($\cdot$) From this, $\operatorname{Cos} B=\frac{1}{2}$ is derived: 2 points.

($\cdot$) A special case with a certain arrangement of radii is considered: 2 points.