# Intersection of circles lie on an angle bisector

Let $$A_1,B_1,C_1$$ be the tangency points of the intersection of the excircles of a triangle $$ABC$$ with the sides $$BC,CA,AB,$$ respectively. Prove that the circumcircles of $$ABB_1$$ and $$ACC_1$$ meet on a bisector of $$\angle BAC.$$

What I thought: Circle $$(ABB_1)$$ has $$\omega_1=b(s-a)$$ and circle $$(ACC_1)$$ has $$v_2=c(s-a)$$. Their radical axis is given by $$-yc(s-a)+zb(s-a)$$ which is equation of $$A$$ angle bisector and we are done.

Let $$D$$ be on $$\overline{BC}$$ so that $$\overline{AD}$$ bisects $$\angle BAC$$. Let $$\varphi$$ be the transformation of the plane formed by composing an inversion with center $$A$$ and radius $$r=\sqrt{AB\cdot AC}$$, with a reflection through $$\overline{AD}$$. Notice that $$\varphi(B)=C$$, $$\varphi(C)=B$$, $$\varphi(\overline{AD})=\overline{AD}$$. Let also $$B_2=\varphi(B_1)$$, $$C_2=\varphi(C_1)$$. Notice that $$\varphi((ABB_1))=\overline{CB_2}$$, $$\varphi((ACC_1))=\overline{BC_2}$$. We now wish to prove that $$\overline{AD}$$, $$\overline{BC_2}$$, $$\overline{CB_2}$$ concur. We do this by Ceva on $$\bigtriangleup ABC$$.
Let $$a=\overline{BC}$$, $$b=\overline{CA}$$, $$c=\overline{AB}$$, $$s=\frac{a+b+c}{2}$$. We have $$\frac{|\overline{AC_2}|}{|\overline{C_2B}|}=\frac{|\overline{AC_2}|}{|\overline{AC_2}|-|\overline{AB}|}=\frac{\frac{r^2}{s-c}}{\frac{r^2}{s-c}-\frac{r^2}{b}}=\frac{b}{b+c-s}.$$ Likewise, $$\frac{|\overline{CB_2}|}{|\overline{B_2A}|}=\frac{|\overline{AB_2}|-|\overline{AC}|}{|\overline{B_2A}|}=\frac{\frac{r^2}{s-b}-\frac{r^2}{c}}{\frac{r^2}{s-b}}=\frac{b+c-s}{c}.$$And finally, $$\frac{|\overline{BD}|}{|\overline{DC}|}=\frac{c}{b},$$by the Bisector Theorem. Cross-multiplying yields what we wanted to prove.