Prove that the distance between the midpoint of side $BC$ of triangle $ABC$ and the midpoint of the arc $ABC$ of its circumscribed circle is not less than $AB / 2$
Apollonius!
We denote the midpoint of arc $ABC$ by $M$ Applying Apollonius theorem on $\triangle MBA$ and $\triangle MBC$,we get that $MD^2+BD^2=MF^2+BF^2$ where $D$ is the midpoint of $BC$ and $F$ is the midpoint of $AB$ .So the problem gets converted and we need to prove that $BD\leq MF$ Let $X$ be a point on $MF$ such that $BXDB$ is a parallelogram It needs to be proven that $X$ lies between $M$ and $F$ Now we consider 3 cases.
(1)$\angle A\leq \angle C$:In that case $\angle XDC=90+\angle B\div 2$ and $\angle FDC=180-\angle C$ Since $90+\angle B\div 2\geq 180-\angle C$, we get that $\angle FDC\leq \angle XDC\Rightarrow M-X-F$
(2)$BA=BC$: In that case we get the equality condition.
(3)$\angle A\geq \angle C$: Working similar to the first case $\blacksquare$
Archimedes!
We donote the midpoint of arc \(ABC\) by\( M \)and that of \(BC\) by \(N\). Note that the projection of \(MN\) on \(BC\) is itself equal to \(AB/ 2\) \(\blacksquare\)