In modern geometry (i.e. Hilbert axioms), how do we show that the area of $\triangle ABC$ is less than the area of the sector centered at $A$? Is there a way to show it without calculus or analytic geometry?
Could we show that the area is less than or equal by saying that if $a$ is an interior point of $\triangle ABC$ then it is an interior point of the sector?
Suppose $AC$ has a length of $1$. Let $\angle CAB=\theta$. Then we have \begin{align} A(\text{triangle})&=\frac12\sin\theta\\ A(\text{sector})&=\frac12\theta \end{align}
You are trying to show the inequality $$\sin\theta\leq \theta$$ which is easy using calculus.
You may also be interested in the circumscribed circle around a regular polygon if you want to take a geometric approach. But I do not believe that you can result on those axioms to make such a statement. If it is shown otherwise, I would gladly redact my answer.