18.7 Let $K$ be the incenter of $\triangle ABC$, and let points $C_{1}$ and $B_{1}$ be the midpoints of sides $AB$ and $AC$, respectively. The line $AC$ intersects $C_{1}K$ at point $B_{2}$, and the line $AB$ intersects $B_{1}K$ at point $C_{2}$. If the area of $\triangle AB_{2}C_{2}$ is equal to the area of $\triangle ABC$, find the size of $\angle CAB$.
18.7 Let the inradius of $\triangle ABC$ be $r$, and the semiperimeter be $p$. The area of $\triangle AC_1B_2 = \frac{1}{2} AC_1 \cdot AB_2 \sin A$; the area of $\triangle ABC = \frac{1}{2} AC \cdot AB \sin A$. Using the area relationship $\triangle AC_1B_2 - \triangle AKB_2 = \triangle AKC_1$, we can derive: $\frac{AB_2 \cdot r p}{2 b} - \frac{AB_2 \cdot r}{2} = \frac{c r}{4}$, from which we can get $AB_2(a-b+c) = bc$ (1); similarly, $AC_2 \cdot (a+b-c) = bc$ (2).
Using the known area relationship, we get $AB_2 \cdot AC_2 = bc$.
Multiplying (1) and (2) yields $(a-b+c)(a+b-c) = bc$, from which we can calculate $\angle CAB = 60^{\circ}$.