Angle bisectors and related circumcircles

Question

The internal and external bisectors of angle $A$ of a triangle $ABC$ meet $BC$ (or the extension of $BC$) at $X$ and $X'$ respectively, show that the circles $ABC$ and $AXX'$ intersect orthogonally. I have shown that the centre of the circle $AXX'$ lies on (line) $BC$, but cannot proceed further. Any advice would be appreciated.

Answer 1

The center of the circle $XAX'$ it's a mid-point of $XX'$ because $$\measuredangle XAX'=90^{\circ}.$$ Let $M$ be this mid-point, $O$ be circumcenter of $\Delta ABC$ and let $AX$ is placed between rays $AO$ and $AM$.

Thus, in the standard notaition: $$\measuredangle OAM=\measuredangle OAX+\measuredangle XAM=\frac{\alpha}{2}-\measuredangle BAO+\measuredangle AXM=$$ $$=\frac{\alpha}{2}-\left(90^{\circ}-\gamma\right)+\beta+\frac{\alpha}{2}=90^{\circ}$$ and we are done!

Answer 2

For two circles $\Gamma$ and $\Omega$ that intersect at $A$ to be orthogonal, it suffices for the line from $A$ to the center of $\Gamma$ be tangent to $\Omega$. Equivalently, if $O$ is the circumcenter of $(ABC)$, we want line $AO$ to be tangent to $(AXX')$.

You can do this with angle-chasing. We have \begin{align*} \angle OAX &=\angle OAB-\angle BAX\\ &=90^\circ-\angle C-\frac{\angle A}{2}\\ &=90^\circ-\angle AXX'\\ &=\angle AX'X, \end{align*} so $AO$ is tangent to $(AXX')$.

Just for fun, here's another proof by inversion: if you invert about $A$ with radius $\sqrt{AB\cdot AC}$ and reflect across the angle bisector (a "root-$BC$ inversion") line $AO$ becomes the $A$-altitude in $\triangle ABC$, while $X$ and $X'$ map to the intersections of the internal and external $A$-angle bisectors with $(ABC)$, which are the midpoints of the two arcs $BC$. The line between these two points is the image of $(AXX')$ in this inversion, so $(AXX')$ simply maps to the perpendicular bisector of $BC$, which is parallel to the $A$-altitude. Since the images of $AO$ and $(AXX')$ are parallel lines, $AO$ and $(AXX')$ are tangent, as desired.