Inscribed angle theorem: half the arc it stands on

Inscribed angle theorem

Why does every angle drawn from the circle onto the same arc come out the same, and exactly half the central angle?

Drag P around the circle: the angle APB stays fixed while P stays on the same side of AB, and it is always half the arc it stands on. Pull AB into a diameter for a right angle, or move P across the chord to get the supplement.

Drag the ringed points on the board.

Middle school
High school
Inscribed angle
Central angle
Angle in a semicircle
Cyclic quadrilateral

\theta_A

3.67

\theta_B

5.76

\theta_P

1.75

\angle APB = \tfrac12\,\overset{\frown}{AB}

$\angle APB$: 60.0
$\overset{\frown}{AB}$: 120.0
$\overset{\frown}{AB} \div \angle APB$: 2.000
$\angle OPA$: 35.0
$\angle OAP$: 35.0
$\angle OPB$: 25.0

∠APB is a right angleno

Guided tour

1Half the arc
The highlighted arc $AB$ is the one $P$ looks at, and it measures $120^\circ$ : the central angle $\angle AOB$ . From $P$ the same arc is seen under $\angle APB=60^\circ$ , exactly half. Both are measured from the coordinates, not set.
2Move P, the angle stays
Drag $P$ anywhere on the upper arc. The chords swing, $\angle OPA$ and $\angle OPB$ change, but $\angle APB$ stays $60^\circ$ : inscribed angles on the same arc are equal.
3Why: one isosceles triangle
Put $P$ opposite $A$ , so $PA$ passes through $O$ . Then $OP=OB$ (both radii), so triangle $OPB$ is isosceles and its base angles are equal. The central angle $\angle AOB$ is the exterior angle at $O$ , so it equals the sum of the two base angles: $2\angle OPB$ . Here $\angle OPB=60^\circ$ and the arc is $120^\circ$ .
4A diameter gives 90°
Make $AB$ a diameter. The arc it cuts off is a semicircle, $180^\circ$ , so every $P$ on the circle sees it under $90^\circ$ .
5Cross the chord
Bring $A$ and $B$ back and drag $P$ onto the lower arc. Now $P$ looks at the other arc, $360^\circ-120^\circ=240^\circ$ , and $\angle APB=120^\circ$ . Together with the $60^\circ$ from above it makes $180^\circ$ : opposite angles of a quadrilateral in a circle are supplementary.

Explanation

1
The theorem
An inscribed angle has its vertex on the circle and its sides along two chords. The arc between its sides, on the far side from the vertex, is the arc it stands on. Measuring an arc by its central angle:
$\angle APB=\tfrac12\,\overset{\frown}{AB}$
Two things follow at once: inscribed angles on the same arc are equal, and an angle on a diameter is $90^\circ$ .
2
Proof from two facts
The proof needs only: base angles of an isosceles triangle are equal, and an exterior angle equals the sum of the two opposite interior angles (both follow from the angle sum $180^\circ$ ).
- ** $O$ on a side.** Say $PA$ passes through $O$ . In triangle $OPB$ , $OP=OB$ , so $\angle OPB=\angle OBP$ . The central angle $\angle AOB$ is exterior at $O$ , so $\angle AOB=2\angle OPB=2\angle APB$ .
- ** $O$ inside the angle.** Draw the diameter $PP'$ (dashed). It splits $\angle APB$ into $\angle OPA+\angle OPB$ and the arc into two pieces. The first case applies to each piece; add them.
- ** $O$ outside the angle.** The same diameter, but now $\angle APB=|\angle OPA-\angle OPB|$ ; subtract instead of add.
On the board, $\angle OPA$ and $\angle OAP$ are measured separately and always agree: that is the isosceles step.
3
The angle in a semicircle
When $AB$ is a diameter the arc is $180^\circ$ , so $\angle APB=90^\circ$ for every $P$ . The converse is how you find a circle's centre with a set square: put the right angle on the rim, mark where the sides cross it, and that chord is a diameter.
4
Quadrilaterals in a circle
Points $P$ and $Q$ on opposite sides of chord $AB$ stand on the two arcs, which add to $360^\circ$ . Halving, $\angle APB+\angle AQB=180^\circ$ : the opposite angles of a cyclic quadrilateral $APBQ$ are supplementary.

Check yourself

Q01

An arc measures $100^\circ$ . What is an inscribed angle standing on it?

Q02

In the proof, why is $\angle OPB=\angle OBP$ ?

Q03

A quadrilateral $ABCD$ is inscribed in a circle and $\angle A=70^\circ$ . What is $\angle C$ ?

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