I know this will seem like an obvious question to many of you, but:
Consider an acute $\triangle ABC$ with $AB=AC$. Also let the midpoint of $BC$ be $M$. Is it always true that $A$, the circumcenter of $\triangle ABC$, and $M$ are collinear?
If it is always true, then please provide a proof. If not, provide a counterexample.
My idea:
Let's say that the circumcenter is $O$. Because $\triangle ABC$ is acute, we know that $O$ is inside the triangle, and therefore $O$ should be between $A$ and $M$. Now, I'm not too sure how I can continue forward with this, but I think we have to prove either $AO+OM=AM$, or $m\angle AOM=180^\circ$, both of which I don't know how to do. I see that $AO$ is the circumradius, so I think we have to find the circumradius in the proof?
Another idea I had was to draw the perpendicular bisectors of each side(they intersect at $O$), and somehow use the surrounding right triangles to find $m\angle AOM$. But I'm not particularly that great at angle chasing and things like that, so I couldn't progress far with this strategy either.
It is well known that the perpendicular bisector of the base in an isosceles triangle passes through the vertex opposite to the base. In order to prove this, we use the same notations as mentioned in the question. Consider the triangles $\triangle ABM, \triangle ACM$, in these two, $AB=AC, BM=CM, AM=AM$. Thus by $SSS$ congruence, these triangles are congruent and hence by CPCT, $\angle AMB=\angle AMC=\frac{\angle AMB+\angle AMC}{2}=90^{\circ}$ which is enough to imply that $AM$ is the perpendicular bisector of $BC$. By the definition of circumcenter, it also lies on the perpendicular bisector of $BC$. Hence $A, O, M$ are collinear where $O$ is the circumcenter of $\triangle ABC$