Let $\triangle$ ABC be a triangle and let $\ell$ be the A-angle bisector. Denote by B' the reflection of B over $\ell$. Prove that the circumcenter of $\triangle$ CIB' lies on $\ell$.
My work: Let D denote the circumcenter of $\triangle$ CIB'. Then it suffices to show that D is also the circumcenter of $\triangle$ BIB' or $\triangle$ BB'C.
Can anyone help me solve this problem? Thanks.
Let $M$ be the point where the angle bisector from $A$ intersects the circumcircle of $ABC$. In other words, $M$ is the midpoint of the circular arc $BC$. I plan to show that $M$ is the circumcenter of triangle $CB'I$.
Indeed, it it easy to see that $MB=MC$ and that $MB=MB'$. It remains to prove that $MB=MI$.
However, it is easy to prove that $\angle IBM =(A + B)/2=(\pi-C)/2$, and $\angle IMB= C$, from which, $\angle MIB = (\pi-C)/2$.
Thus, triangle $IBM$ is isosceles, and therefore, $MB=MI$.
We have, $MB=MB'=MC=MI$, which means $M$ is the circumcenter of $BCB'I$ as promised.