9. (HUN 3) The incenter of the triangle $A B C$ is $K$. The midpoint of $A B$ is $C_{1}$ and that of $A C$ is $B_{1}$. The lines $C_{1} K$ and $A C$ meet at $B_{2}$, the lines $B_{1} K$ and $A B$ at $C_{2}$. If the areas of the triangles $A B_{2} C_{2}$ and $A B C$ are equal, what is the measure of angle $\angle C A B$?
9. Let $a, b, c$ be the lengths of the sides of $\triangle ABC, s=\frac{a+b+c}{2}, r$ the inradius of the triangle, and $c_{1}$ and $b_{1}$ the lengths of $AB_{2}$ and $AC_{2}$ respectively. As usual we will denote by $S(XYZ)$ the area of $\triangle XYZ$. We have  $$ \begin{gathered} S\left(AC_{1}B_{2}\right)=\frac{AC_{1} \cdot AB_{2}}{AC \cdot AB} S(ABC)=\frac{c_{1} r s}{2 b} \\ S\left(AKB_{2}\right)=\frac{c_{1} r}{2}, \quad S\left(AC_{1}K\right)=\frac{c r}{4} \end{gathered} $$  From $S\left(AC_{1}B_{2}\right)=S\left(AKB_{2}\right)+S\left(AC_{1}K\right)$ we get $\frac{c_{1} r s}{2 b}=\frac{c_{1} r}{2}+\frac{c r}{4}$; therefore $(a-b+c) c_{1}=b c$. By looking at the area of $\triangle AB_{1}C_{2}$ we similarly obtain $(a+b-c) b_{1}=b c$. From these two equations and from $S(ABC)=S\left(AB_{2}C_{2}\right)$, from which we have $b_{1} c_{1}=b c$, we obtain  $$ a^{2}-(b-c)^{2}=b c \Rightarrow \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\cos (\angle BAC)=\frac{1}{2} \Rightarrow \angle BAC=60^{\circ} $$