5. Let $K$ be the incenter of $\triangle A B C$, and let 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 $S_{\triangle A B_{2} c_{2}}=S_{\triangle A B C}$, find $\angle C A B$.
5. Let $BC=a, CA=b, AB=c, AC_{2}=b^{\prime}, AB_{2}=c^{\prime}, p=\frac{1}{2}(a+b+c)$, and let the inradius of $\triangle ABC$ be $r$. From $S_{\triangle AC_{1}B_{2}}=\frac{1}{2} AC_{1} \cdot AB_{2} \cdot \sin A, S_{\triangle ABC}=\frac{1}{2} AC \cdot AB \cdot \sin A, C_{1}$ is the midpoint of $AB$ and $S_{\triangle ABC}=rp$, we can deduce that $S_{\triangle AC_{1}B_{2}}=\frac{AC_{1} \cdot AB_{2}}{AB \cdot AC} \cdot S_{\triangle ABC}=\frac{c^{\prime} rp}{2b}$. But $S_{\triangle AKB_{2}}=\frac{1}{2} c^{\prime} r, S_{\triangle AKC_{1}}=\frac{1}{4} cr$, noting that $S_{\triangle AC_{1}B_{2}}-S_{\triangle AKB_{2}}=S_{\triangle AKC_{1}}$, then $\frac{c^{\prime} rp}{2b}-\frac{c^{\prime} r}{2}=\frac{cr}{4}$, hence $2c^{\prime}(p-b)=bc$, i.e., $(a-b+c)c^{\prime}=bc$, similarly we get $(a+b-c)b^{\prime}=bc$. On the other hand, from the known $S_{\triangle AB_{2}C_{2}}=S_{\triangle ABC}$ we get $b^{\prime}c^{\prime}=bc$. Thus, from $(a-b+c)(a+b-c)b^{\prime}c^{\prime}=b^{2}c^{2}$ we get $a^{2}-(b-c)^{2}=bc$, so $a^{2}=b^{2}+c^{2}-bc$. By the cosine rule, $\cos \angle CAB=\frac{1}{2}$, hence $\angle CAB=60^{\circ}$.