Inscribed angle

An inscribed angle $\angle APB$ has its vertex P on a circle and its sides through two other points A, B of the circle. It stands on the arc from A to B that does not contain P, and the central angle on that arc is the angle at the centre O between OA and OB — the reflex angle when the arc is more than a half circle. The inscribed angle theorem says the inscribed angle is half the central angle on the same arc. So every inscribed angle on one arc is the same, an angle in a semicircle is a right angle (Thales), and opposite angles of a cyclic quadrilateral add up to 180°.

Isosceles draws the proof. OP, OA and OB are radii, so the triangles OPA and OPB are isosceles, with base angles $\alpha = \angle OPA$ and $\beta = \angle OPB$. Extend PO to the far end P′ of the diameter: the exterior angles at O are 2α and 2β. When O is inside the angle they make up the central angle and the inscribed angle is $\alpha + \beta$; when O is outside, the central angle is their difference and the inscribed angle $|\alpha - \beta|$. As P moves, α and β trade off while their sum or difference stays put; the readout switches from + to − exactly where P′ passes A or B and O lies on a side of the angle.

The page computes α and β from P's own position and sets ∠APB beside “arc ÷ 2”: two numbers worked out separately that always agree. Positions on the circle are held exactly, in thousandths of a degree, and a drag or a wheel moves them in steps of 0.1° (half a pixel on screen), so every angle shown with “=” is the exact value at that point. With P off the circle the angle is measured and marked ≈, except at O, where it is the central angle exactly.