Isosceles geometry contest question

Question

Given a triangle $\triangle ABC$, $\overline{AC} = \overline{BC} > \overline{AB}$. Point O is the center of the circle circumscribed by triangle $\triangle ABC$. Point $M$ is the midpoint of side $\overline{AC}$.

The circumscribed circle of the triangle $\triangle AMO$ intersects the segment $\overline{BM}$ at the point $X$ different from $M$. Prove that $\overline{CX} = 2\overline{MX}$.

I drew it in geogebra and found that $BCOX$ is cyclic, but I have no idea how to prove it and use it. Any hints?

Answer 1

Observations towards a solution. Fill in the gaps as needed.

• [Validating OP claim] $BCOX$ is concyclic
  • By angle chasing, $ \angle OXM = \angle OAM = \angle OCA = \angle OCB = \frac{ \gamma}{2}$.
• [Some Wishful thinking to provide further direction] If $\frac{ CX}{MX} = 2$, what other ratios of 2 are there?
  • $\frac{ CA}{CM} = 2$ should come to mind.
  • This implies that $XA$ is the exterior angle bisector of $CXM$.
• We now show that $XA$ is the exterior angle bisector of $CXM$ to complete the solution.

  • $BCOX$ is concyclic so $\angle CXO = \angle CBO = \frac{\gamma}{2}$.

  • $\angle OXM = \frac{\gamma}{2}$, so $OX$ is the internal angle bisector of $CXM$.

  • $OMAX$ is concyclic so $\angle OXA = 90^\circ$.

  • Hence $XA$ is the exterior angle bisector of $CXM$.