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← Plane geometry IChapter III · Circles: tangents, inscribed angles, concyclicity29 / 50

Lesson · Inscribed angle

Inscribed angles

One theorem carries most of circle geometry: the angle at which a chord is seen from the circle does not depend on where on the arc you stand. Read forwards it computes angles; read backwards it produces circles that were never drawn.

Central and inscribed

Theorem

Let \(AB\) be a chord of a circle with centre \(O\), and let \(P\) be a point of the arc on one side of \(AB\). Then

\[ \angle APB = \tfrac{1}{2}\,\angle AOB , \]

where \(\angle AOB\) is the central angle over the arc not containing \(P\). In particular all points of that arc see \(AB\) under the same angle.

Three consequences are used constantly:

  • Thales. If \(AB\) is a diameter, the central angle is \(180^\circ\), so every point of the circle sees \(AB\) at \(90^\circ\) - and conversely, a point seeing \(AB\) at \(90^\circ\) lies on the circle with diameter \(AB\).
  • Same arc, same angle. \(\angle APB = \angle AQB\) whenever \(P\), \(Q\) are on the same side of \(AB\).
  • Opposite sides, supplementary angles. If \(P\) and \(Q\) are on opposite arcs, then \(\angle APB + \angle AQB = 180^\circ\) - the cyclic quadrilateral rule.

The theorem as a locus

Principle

Fix a segment \(AB\) and an angle \(\varphi\). The set of points \(P\) with \(\angle APB = \varphi\) consists of two circular arcs over \(AB\), one on each side, symmetric in the line \(AB\). So a hypothesis "\(P\) sees \(AB\) under \(\varphi\)" is not a vague constraint: it puts \(P\) on a specific circle you can draw.

AB PQ
\(P\) and \(Q\) are anywhere on the upper arc: the two marked angles are equal.

Example

Two chords \(AB\) and \(CD\) of a circle meet at an interior point \(P\). Then \(\angle PAC = \angle PDB\) (they subtend the arc \(BC\)) and the vertical angles at \(P\) are equal, so the triangles \(APC\) and \(DPB\) are similar. Comparing the corresponding sides gives \(PA \cdot PB = PC \cdot PD\).

Warning

"Same angle" and "supplementary angles" swap places depending on which side of the chord the two points are. Before you write \(\angle APB = \angle AQB\), check that \(P\) and \(Q\) really are on the same arc; if the configuration can vary, say which case you are treating.

Warm-up

Example

Let \(ABC\) be a triangle inscribed in a circle \(k\), and let \(M\) be the midpoint of the arc \(BC\) that does not contain \(A\). Prove that \(AM\) bisects the angle \(\angle BAC\) and that \(MB = MC\).

What does "midpoint of an arc" give you?

The two arcs \(BM\) and \(MC\) are equal. Equal arcs are subtended by equal inscribed angles - and also by equal chords.

Look at the inscribed angles from \(A\)

\(\angle BAM\) stands on the arc \(BM\) and \(\angle MAC\) on the arc \(MC\). Equal arcs, equal angles - so \(AM\) is the bisector.

And the chords

Equal arcs in one circle are cut off by equal chords, so \(MB = MC\): the arc midpoint sits on the perpendicular bisector of \(BC\) as well as on the bisector from \(A\). That is the standard "the bisector from \(A\) meets the circumcircle at the arc midpoint" lemma - remember it.

← Previous · Warm-upA circle is inscribed in triangle \(ABC\), whose sides have lengths \(BC = a\), \(CA = b\) and \(AB = c\). A line tangent to this circle meets the side \(AC\) at the point \(P\) and the side \(BC\) at …Next · Practice →In a quadrilateral \(ABCD\), \[ \angle ABC = 104^\circ, \qquad \angle ADC = 128^\circ, \qquad AB = BC = 2. \] Compute the length of the diagonal \(BD\).