Let $\omega_a$ be the $A$-excircle of the triangle $ABC$, and $A(.)$ is the circle centred at $A$ with zero-radius. We aim to show that $B_2C_2$ is the radical axis of these two circles, and this would be enough to arrive at the result stated in the problem. The following lemma helps us in this matter. Lemma. Let $ABC$ be a triangle; points $X$ and $Y$ are chosen on the sides $AB$ and $AC$ such that $BX=CY$. Let $M$ and $N$ be the midpoints of segments $BY$ and $CX$, respectively. Then $MN$ is perpendicular to the (internal) bisector-line of $\angle BAC$. Proof of the lemma. We know that $\overline{MN} = \overline{BX}+\overline{YC}$. If we denote the unit vector on the direction of the bisector of $\angle BAC$ by $\overline{u}$, then this vector makes equal angles with vectors $\overline{XB}$ and $\overline{YC}$. Since $BX=CY$, we obtain $\overline{XB}\cdot \overline{u} = \overline{YC} \cdot \overline{u}$, where $\cdot$ denotes the usual dot product. Hence $\overline{BX}\cdot \overline{u} + \overline{YC} \cdot \overline{u} = \overline{0}$ and $\overline{MN}\perp \overline{u}$. Moving back to our problem, we see that $BC_1=CB_1$ (as they both are equal to $s-a$, where $s$ is the semi-perimeter), so we can apply the lemma. Assume that $B_2C_2$ meets the sides $AC$ and $AB$ at points $T$ and $S$, respectively. Let $C_1'$ be the symmetrical point to $C_1$ with respect to the bisector of $\angle BAC$. It is not difficult to see that $C_1'$ lies on the side $AC$ and $AC_1'=AC_1=s-b$. Moreover, $C_2T$ should be the midline of the triangle $CC_1C_1'$ parallel to the side $C_1C_1'$. Then \[ AT = AC_1' + \frac{CC_1'}{2} = s-b + \frac{b- (s-b)}{2} = \frac{s}{2}. \]As $S$ and $T$ are symmetrical with respect to the bisector of $\angle BAC$, we get $AS=AT=\frac{s}{2}$. Now, using $\mathrm{Pow}_{A} (\omega_a) = s$, we finally get that \[ \mathrm{Pow}_S (\omega_a) = \mathrm{Pow}_T (\omega_a)=\mathrm{Pow}_S (A(.)) = \mathrm{Pow}_T (A(.)) = \frac{s}{2}, \]i.e. $S$ and $T$ lie on the radical axis of the circles $\omega_a$ and $A(.)$; hence, $W$ also lies.

Let $B_2C_2$ intersects the sides at point M and N.From Menelaus theorem in triangle $BB_1C$ and the collinear points $W,C_2,N$.We know that $\frac{WB}{WC} \cdot \frac{BC_2}{C_2B_1} \cdot \frac {B_1N}{NC}=1$ then $\frac{WB}{WC} \cdot \frac {B_1N}{NC}=1$ and in triangle $BC_1C$ menelaus theorem we find $\frac{WB}{WC} \cdot \frac {BM}{MC_1}=1.$ Then $\frac {B_1N}{NC}=\frac {BM}{MC_1}.$ We know $BC_1=CB_1$, from these equality $CN=C_1M$ and $BM=NB_1$.From menelaus theorem in triangle ABC and for points $W,M,N$ $ we know \frac{WB}{WC} \cdot \frac{BM}{MA} \cdot \frac {AN}{NC}=\frac{WB}{WC} \cdot \frac{B_1N}{NC} \cdot \frac{AM}{AN}=1.$ It gives $AM=AN$.Let A-excircle tangent to sides AB and AC at X and Y.We know $AB_1=BX$.$AM=AN=AB_1+B_1N=BX+BM=XM $ then M is midpoint of AX and similarly N midpoint of AY .Then MN radical axis for A-excircle and A.

@mathuz Nice solution

A solution similar to @mathuz, but slightly easier in my opinion.