Angles in the same segment are equal

Statement

Let $A$, $B$, $C$ and $D$ be distinct concyclic points of the Euclidean plane $\mathbb{R}^2$. Let $\measuredangle$ be the oriented angle, with values in $\mathbb{R} / 2\pi\mathbb{Z}$. Then $$2 \, \measuredangle ACB = 2 \, \measuredangle ADB.$$ For the unoriented angles: if $C$ and $D$ are on the same side of the line $AB$, then $\angle ACB = \angle ADB$. If they are on opposite sides, then $\angle ACB + \angle ADB = \pi$.

Remarks

The statement follows from the inscribed angle theorem: both sides are equal to the oriented angle $\measuredangle AOB$ at the centre $O$. The factor $2$ is necessary. It removes the difference of $\pi$ between points on the two arcs. The case of opposite sides says that the opposite angles of a convex cyclic quadrilateral add up to $\pi$.

Explanations

L1 human

A segment of a circle is the region between a chord $AB$ and one of its arcs. From every point of that arc, the chord $AB$ is seen under the same angle. From the other arc, it is seen under the supplementary angle.