7. (Hungary 3) Let $k$ be the incenter of $\triangle A B C$, $G_{1}, B_{1}$ be the midpoints of sides $A B, A C$ respectively, and let $A C$ intersect $C_{1} K$ at point $B_{2}$, and line $A B$ intersect $B_{1} K$ 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 $\angle C A B$.
Solution 1: Let $a=BC, b=CA, c=AB$, $b^{*}=AC_{2}, \quad c^{*}=AB_{2}, \quad s=\frac{1}{2}(a+b+c)$.
Let $r$ be the inradius of $\triangle ABC$. Since
$$
\begin{array}{l}
S_{\triangle C_{1} B_{2}}=\frac{1}{2} AC_{1} \cdot AB_{2} \sin \angle A, \\
S_{\triangle B C}=\frac{1}{2} AC \cdot AB \sin \angle A,
\end{array}
$$

thus, by $C_{1}$ being the midpoint of $AB$ and taking $S_{\triangle ABC}=r_{3}$, we can deduce that
$$
S_{\triangle A C_{1} B_{2}}=\frac{AC_{1}}{AB} \div \frac{AB_{2}}{AC} S_{\triangle B O}=\frac{c^{*} r s}{2 b} .
$$

But

noting that
$$
S_{\triangle C_{1} B_{2}}-S_{\triangle I_{2}}=S_{\triangle A T O_{1}},
$$

thus,
$$
\begin{array}{l}
\frac{c^{*} r s}{2 b}-\frac{c^{*} r}{2} \\
=\frac{c r}{4} .
\end{array}
$$

Therefore, $2 c^{*}(s-b)$
$$
=bc,
$$

which means $(a-b+c) c^{*}=bc$.
Similarly, we can get $(a+b-c) b^{*}=bc$.
Furthermore, from the given condition $S_{\triangle A R C_{2}}=S_{\triangle A O}$, we have $b^{*} c^{*}=bc$. Therefore,
$$
(a-b+c)(a+b-c) b^{*} c^{*}=b^{2} c^{2}
$$

which simplifies to
$$
a^{2}-(b-c)^{2}=bc .
$$

Thus, $a^{2}=b^{2}+c^{2}-bc$. From the cosine rule, we know $\cos \angle CAB=\frac{1}{2}$, hence $\angle CAB=60^{\circ}$.

Solution 2: Let $E$ be the point where the incircle of $\triangle ABC$ touches $AB$, $E^{*}$ be the point symmetric to $E$ with respect to $K$, and $F$ be the point symmetric to $E$ with respect to $C_{1}$ (Figure 4).
By the well-known

conclusion about $C$,
$E^{*}$, and $F$ being
collinear,
and noting
that $KC_{1}$ is
the midline of
$\triangle EFE^{*}$, we know $CF \| B_{2}C_{1}$. Therefore,
$$
\frac{AB_{2}}{AC_{1}}=\frac{AC}{AF} .
$$

Similarly, if we assume $G$ is the point on side $AC$ symmetric to the tangency point with respect to $B$, then
$$
\frac{AC_{2}}{AB_{1}}=\frac{AB}{AG} .
$$

Thus,
$$
\frac{AB_{2} \cdot AC_{2}}{AB \cdot AC}=\frac{AB_{1} \cdot AC_{1}}{AG \cdot AF} .
$$

But the left side of the above equation equals: the ratio of the areas of $\triangle AB_{2}C_{2}$ to $\triangle ABC$, which, by the given condition, is 1. Therefore,
$$
AB_{1} \cdot AC_{1}=AG \cdot AF .
$$

Using the notations $a, b, c$ as in Solution 1, and noting that $AB_{1}=\frac{b}{2}, AC_{1}=\frac{c}{2}$, and
$$
AF=\frac{a+c-b}{2}, AG=\frac{a+b-c}{2} .
$$

Substituting these into the previous equation, we get
$$
a^{2}=b^{2}+c^{2}-bc .
$$

Thus, $\angle CAB=60^{\circ}$.