一、(50 分) Let $K$ be the incenter of $\triangle A B C$, and points $C_{1}, B_{1}$ be the midpoints of sides $A B, A C$ respectively. Line $A C$ intersects $C_{1} K$ at point $B_{2}$, and line $A B$ intersects $B_{1} K$ at point $C_{2}$. If the area of $\triangle A B_{2} C_{2}$ is equal to the area of $\triangle A B C$, find $\angle C A B$.

保留源文本的换行和格式，直接输出翻译结果如下：

一、(50 分) Let $K$ be the incenter of $\triangle A B C$, and points $C_{1}, B_{1}$ be the midpoints of sides $A B, A C$ respectively. Line $A C$ intersects $C_{1} K$ at point $B_{2}$, and line $A B$ intersects $B_{1} K$ at point $C_{2}$. If the area of $\triangle A B_{2} C_{2}$ is equal to the area of $\triangle A B C$, find $\angle C A B$.
Let $a=BC, b=CA, c=AB, b^{*}=AC_{2}, c^{*}=AB_{2}, s=\frac{1}{2}(a+b+\epsilon)$. Suppose the inradius of $\triangle ABC$ is $r$.
$\because S_{\triangle AC_{1}B_{2}}=\frac{1}{2} AC_{1} \cdot AB_{2} \sin A$,
$S_{\triangle ABC}=\frac{1}{2} AC \cdot AB \sin A$,
$\therefore$ From $C_{1}$ being the midpoint of $AB$ and $S_{\triangle ABC}=rs$, we can deduce that
$S_{\triangle AC_{1}B_{2}}=\frac{AC_{1} \cdot AB_{2}}{AB \cdot AC} \cdot S_{\triangle ABC}=\frac{c^{*} rs}{2b}$.
But $S_{\triangle AKB_{2}}=\frac{c^{*} r}{2}, S_{\triangle AKC_{1}}=\frac{cr}{4}$,
Noting that $S_{\triangle AC_{1}B_{2}}-S_{\triangle AKB_{2}}=S_{\triangle AKC_{1}}$,
then $\frac{c^{*} rs}{2b}-\frac{c^{*} r}{2}=\frac{cr}{4}$.
fir $2c^{*}(s-b)=bc$,

Thus $(a+b-c)c^{*}=bc$.
Similarly, we can get $(a+b-c)b^{*}=bc$.
On the other hand, from the known $S_{\triangle AB_{2}C_{2}}=S_{\triangle ABC}$, we get $b^{*}c^{*}=bc$. Therefore, from $(a-b+c)(a+b-c)b^{*}c^{*}=b^{2}c^{2}$, we have $a^{2}-(b-c)^{2}=bc$.
Hence $a^{2}=b^{2}+c^{2}-bc$.
Thus, by the cosine rule, $\cos \angle CAB=\frac{1}{2}$,
Therefore, $\angle CAB=60^{\circ}$.