A circle has center O. A chord AB joins two points on the circle, and you can drag A and B around the circle. From the center O, a perpendicular is dropped to the chord, meeting it at point M. The theorem: this perpendicular always meets the chord at its midpoint, so AM equals MB. The right angle at M and the two equal halves are marked. Together they form a right triangle OMA, so the radius, the distance from center to chord, and half the chord satisfy r squared equals OM squared plus half-AB squared. The half-chord, the distance OM, and the check that AM equals MB are shown live. An info button opens a drawer with the explanation.