Let points $B_1$ and $C_1$ be on the sides, and $A_1$ is on the extension of the side $BC$ in direction of point $B$. By the Menelaus' theorem
$$\frac{CB_2 \ AC_2 \ BA_2}{B_2A \ C_2B \ A_2C} = \frac{AB_1 \ BC_1 \ CA_1}{B_1C \ C_1A \ A_1B} = 1,$$
so points $A_2$, $B_2$ and $C_2$ are collinear (fig. 25).

Let $M$ be the middle of $AB$, $N$ be the middle of $AC$, $K$ be the base of the perpendicular dropped from a point $I$, the center of the inscribed circle, to a straight line $B_2C_2$. Triangles $C_2IM$ and $C_1IM$ are equal, so $C_2I = IC_1$, $\angle C_2IC_1 = 180^\circ - 2\angle B_1C_1A$. Similarly, $B_2I = B_1I$ and $\angle B_2IB_1 = 180^\circ - 2\angle C_1B_1A$. Then,
$$\angle B_2IC_2 = 2\angle B_1C_1A + 2\angle C_1B_1A - 180^\circ = 2(180^\circ - \angle A) - 180^\circ = 60^\circ.$$
Since $\angle B_2IC_2 = \angle B_1AC_1$ and $\frac{B_2I}{C_2I} = \frac{B_1I}{C_1I} = \frac{B_1A}{C_1A}$ ($I$ is a base of the bisectrix $AI$ of the triangle $B_1AC_1$), triangles $B_2IC_2$ and $B_1AC_1$ are homothetic, hence $\angle C_2B_2I = \angle AB_1C_1$ and $\angle B_2C_2I = \angle AC_1B_1$. The segment $IB_2$ is common for the right-angled triangles $B_2MI$ and $B_2KI$, and also $\angle NB_2I = \angle NB_1I = \angle KB_2I$, so $NB_2 = KB_2$. In the same way, $KC_2 = C_2M$. Since
$$B_2C + C_2B = B_2N + NC + BM + MC_2 = B_2K + KC_2 + BC = B_2C_2 + BC,$$
the quadrangle $CB_2C_2B$ is circumscribed, therefore the straight line $B_2C_2$ touches the inscribed circle of the triangle $ABC$.