The inscribed angle is half of the corresponding central angle
Theorem.
Let $A$, $B$, and $P$ be distinct points on a circle with center $O$.
Let $\angle AOB$ denote the central angle, possibly reflex, subtending
the arc $AB$ that does not contain $P$. Then
$$\angle APB \;=\; \tfrac{1}{2}\,\angle AOB.$$
In particular, if $P$ and $Q$ lie on the same arc from $A$ to $B$, then
$\angle APB=\angle AQB$; and if $AB$ is a diameter, then
$\angle APB=90^\circ$.
Drag the points $A$, $B$, $P$ (and $Q$)
The highlighted arc is the arc $AB$ not containing $P$; it is intercepted
by $\angle APB$ and subtended at the center by $\angle AOB$.
Live values
Corresponding central angle $\angle AOB$ (possibly reflex)
—
Inscribed angle $\angle APB$
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Inscribed angle $\angle AQB$
—
Ratio $\angle AOB / \angle APB$
—
When $P$ passes from one arc to the other, the angle is momentarily undefined
at $A$ or $B$. Once $P$ is distinct from $A$ and $B$ again, the highlighted arc
switches, and the identity $\angle APB = \tfrac12 \angle AOB$ continues to hold;
inscribed angles on the two opposite arcs are supplementary.