Simplest proof that the edge of an inscribed equilateral triangle bisects the radius

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Context: I am giving a short talk on the Bertrand Paradox to a mixed group, many of whom have studied mathematics at a higher level some years ago. The point of the talk is the philosophical ramifications on the Principle of Indifference, but I would very much like to be able to prove all of the results I'm relying on from first principles as my audience are a pedantic bunch.

Question:

Given a circle with an inscribed equilateral triangle and a radius that is perpendicular to one of the triangle's sides, I need to be able to show that the radius is bisected by the triangle (not that a perpendicular radius bisects a chord!).

This is fairly simple by route of the Central Angle Theorem and then the 30-60-90 triangle, but that gets rather involved if I have to prove even a limited version of the Central Angle Theorem. Is there a more direct route I'm not seeing?

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The centroid trisects the median.

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