HG10 - Angles in circles | Eduqas GCSE Mathematics

Learn to prove and use angle theorems in a circle, then connect them in reasoned angle chains. HG10 requires the geometry that justifies an answer as well as its numerical value.

Centre and circumference

If you move a point along one arc of a circle, can the angle it makes with a fixed chord stay unchanged? The answer depends on which side of the chord the point stays.

A chord joins two points on a circle. An arc is a portion of its circumference. Two distinct endpoints $A,B$ divide the circumference into two arcs. For a third circle point $C$ , angle $ACB$ stands on the arc $AB$ that does not contain $C$ . The corresponding central angle follows that same arc; it may be reflex.

Centre and circumference theorem

The angle at the centre is twice the angle at the circumference standing on the same arc:

\text{selected central angle }AOB=2\angle ACB.

Here $O$ is the centre and $A,B,C$ are distinct points on the circle. Choose the central angle belonging to the arc away from $C$ , rather than automatically choosing the smaller angle.

Why it is true. Join $OA,OB,OC$ , and extend $CO$ through $O$ to a second circle point $D$ . First suppose neither $A$ nor $B$ is $D$ . In isosceles triangle $AOC$ , let $\angle ACO=\alpha$ . Its other base angle is also $\alpha$ , so the external angle $AOD$ is $2\alpha$ . Similarly, writing $\angle BCO=\beta$ gives external angle $BOD=2\beta$ .

If ray $CO$ lies inside $\angle ACB$ , add the two pieces: the circumference angle is $\alpha+\beta$ and the central angle along the arc away from $C$ is $2\alpha+2\beta$ . This central angle can be reflex. If $CO$ lies outside, subtract the smaller piece from the larger at both vertices. Again the central angle is twice the circumference angle. If one arm $CA$ or $CB$ is the diameter $CD$ , the one remaining isosceles triangle gives the result directly by its external angle.

The auxiliary diameter lets the same proof cover both positions of the centre.

[DIAGRAM: asset_name: Proving the centre theorem; asset_slug: centre-theorem-proof; caption: Equal radii make isosceles triangles; add or subtract their external angles.; description: Two exact unit-circle configurations join C through O to its opposite point D and join A and B to C and O. In the first, CO is inside angle ACB and external angles at O add; in the second CO lies outside angle ACB and the external angles subtract. All named points and radii are explicit.]
Equal radii make isosceles triangles

Angles in the same segment now follow. If $C$ and $D$ lie on the same arc between chord endpoints $A,B$ , both $\angle ACB$ and $\angle ADB$ are half the same central angle. Therefore they are equal. The vertices must lie on the same side of chord $AB$ ; merely sharing the chord is insufficient.

Worked example 1: a non-reflex central angle

Question. $O$ is the centre. The minor central angle $AOB$ is $136^\circ$ . Points $C,D$ lie on the major arc $AB$ . Find $\angle ACB$ and $\angle ADB$ .

Plan. Both circumference vertices are away from the minor arc, so use the given minor central angle, then the same-segment theorem.

\angle ACB=\frac{136^\circ}{2}=68^\circ, \qquad \angle ADB=68^\circ.

Answer. Both angles are $68^\circ$ .

Check. Doubling either gives the selected $136^\circ$ . The alternative central angle $224^\circ$ belongs to the arc containing $C,D$ , so it is not the angle for these vertices.

Change the selected arc to a major arc and put both circumference vertices on the other, minor arc. The central angle is now reflex.

Worked example 2: a reflex central angle

Question. The reflex central angle $AOB$ is $248^\circ$ . Points $C,D$ lie on the minor arc $AB$ . Find $\angle ACB$ and $\angle ADB$ .

Plan. The arc away from $C,D$ is the major arc, so halve the reflex angle.

\angle ACB=\frac{248^\circ}{2}=124^\circ, \qquad \angle ADB=124^\circ.

Answer. Both angles are $124^\circ$ .

Check. Doubling gives $248^\circ$ , and an obtuse circumference angle is consistent with standing on a major arc. Halving the other central angle $112^\circ$ would give the angle for a vertex on the other side of the chord.

The point positions determine which central angle belongs to the theorem.

[DIAGRAM: asset_name: Minor and reflex central angles; asset_slug: selected-circle-arcs; caption: Both central and circumference angles must stand on the same selected arc.; description: Exact unit-circle panels show central angles136 and reflex248 with two circumference vertices on the opposite arc. Their corresponding inscribed angles are68 and124. The selected arc is highlighted, with centre O and all endpoint labels.]
Both central and circumference angles must stand on the same selected arc.

Watch out

Do not halve a central number before checking the arc. Do not claim two angles are equal just because they stand on the same chord: check that their vertices lie in the same segment.

Use a fresh reflex angle.

Semicircles and cyclic quadrilaterals

Two useful results follow from the centre theorem. Their hypotheses are essential: a chord drawn almost through the centre is not a diameter, and an arbitrary quadrilateral is not cyclic.

Angle in a semicircle

If $AC$ is a diameter and $B$ is any other point on the circle, then $\angle ABC=90^\circ$ .

Proof. The central angle standing on either semicircular arc $AC$ is $180^\circ$ . The circumference angle is half of it: $180^\circ/2=90^\circ$ . The right angle is at $B$ , opposite the diameter, not at an arbitrary circle point or at an endpoint of the diameter.

A cyclic quadrilateral has four distinct vertices on one circle, named in order around its circumference.

Opposite angles of a cyclic quadrilateral

For cyclic quadrilateral $ABCD$ ,

\angle BAD+\angle BCD=180^\circ, \qquad \angle ABC+\angle ADC=180^\circ.

Proof. Opposite vertices $A,C$ stand on the two complementary arcs with endpoints $B,D$ . Their corresponding central angles together make a full turn of $360^\circ$ . Each circumference angle is half its own central angle, so their sum is $180^\circ$ . Repeat for the other opposite pair. Consequently an exterior angle of a cyclic quadrilateral equals the opposite interior angle: both supplement the same interior angle.

Now combine a diameter, a triangle angle sum and a cyclic angle sum in one configuration.

Worked example 3: diameter and cyclic angles

Question. $A,B,C,D$ lie in that order around a circle. $AC$ is a diameter, $\angle BAC=34^\circ$ , and $\angle CAD=22^\circ$ . Find $\angle ABC$ , $\angle BCA$ and $\angle BCD$ .

Plan. Use the diameter for the first right angle; find the triangle angle and the whole angle at $A$ before using cyclic opposites.

\begin{aligned} \angle ABC&=90^\circ &&\text{angle in a semicircle},\\ \angle BCA&=180^\circ-90^\circ-34^\circ=56^\circ &&\text{triangle angle sum},\\ \angle BAD&=34^\circ+22^\circ=56^\circ,\\ \angle BCD&=180^\circ-56^\circ=124^\circ &&\text{opposite cyclic angles}. \end{aligned}

Answer. $\angle ABC=90^\circ$ , $\angle BCA=56^\circ$ , $\angle BCD=124^\circ$ .

Check. Triangle $ACD$ also has $\angle ADC=90^\circ$ , hence $\angle ACD=68^\circ$ . The whole angle at $C$ is $56^\circ+68^\circ=124^\circ$ , agreeing by a different route.

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HG10 - Angles in circles

Centre and circumference

Semicircles and cyclic quadrilaterals

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