# 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.

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 $AB{B}_{1}$ and $AC{C}_{1}$ meet on a bisector of $\mathrm{\angle }BAC.$
What I thought: Circle $\left(AB{B}_{1}\right)$ has ${\omega }_{1}=b\left(s-a\right)$ and circle $\left(AC{C}_{1}\right)$ has ${v}_{2}=c\left(s-a\right)$. Their radical axis is given by $-yc\left(s-a\right)+zb\left(s-a\right)$ which is equation of A angle bisector and we are done.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Alannah Yang
Let $D$ be on $\overline{BC}$ so that $\overline{AD}$ bisects $\mathrm{\angle }BAC$. Let $\phi$ 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 $\phi \left(B\right)=C$, $\phi \left(C\right)=B$, $\phi \left(\overline{AD}\right)=\overline{AD}$. Let also ${B}_{2}=\phi \left({B}_{1}\right)$, ${C}_{2}=\phi \left({C}_{1}\right)$. Notice that $\phi \left(\left(AB{B}_{1}\right)\right)=\overline{C{B}_{2}}$, $\phi \left(\left(AC{C}_{1}\right)\right)=\overline{B{C}_{2}}$. We now wish to prove that $\overline{AD}$, $\overline{B{C}_{2}}$, $\overline{C{B}_{2}}$ concur. We do this by Ceva on $△ABC$.

Let $a=\overline{BC}$, $b=\overline{CA}$, $c=\overline{AB}$, $s=\frac{a+b+c}{2}$. We have
$\frac{|\overline{A{C}_{2}}|}{|\overline{{C}_{2}B}|}=\frac{|\overline{A{C}_{2}}|}{|\overline{A{C}_{2}}|-|\overline{AB}|}=\frac{\frac{{r}^{2}}{s-c}}{\frac{{r}^{2}}{s-c}-\frac{{r}^{2}}{b}}=\frac{b}{b+c-s}.$
Likewise,
$\frac{|\overline{C{B}_{2}}|}{|\overline{{B}_{2}A}|}=\frac{|\overline{A{B}_{2}}|-|\overline{AC}|}{|\overline{{B}_{2}A}|}=\frac{\frac{{r}^{2}}{s-b}-\frac{{r}^{2}}{c}}{\frac{{r}^{2}}{s-b}}=\frac{b+c-s}{c}.$
And finally,
$\frac{|\overline{BD}|}{|\overline{DC}|}=\frac{c}{b},$
by the Bisector Theorem. Cross-multiplying yields what we wanted to prove.