International Baccalaureate · MYP 5

Circle Theorems and Geometric Proof

⌚ ~106 min ▢ 11 blocks ✓ 44 exercises ➣ Prerequisites : Similarity and Congruence, Deductive Geometry and Circle Theorems, Right-Angled Trigonometry and Bearings, Quadratic Equations

I Angles in a Circle

I.1 Angle at the Centre

Discover

In an open-air theatre, the front of the stage is the segment $\Segment{AB}$, and the seats are placed on the arc of a circle of centre $O$ through $A$ and $B$ drawn in bold. Two spectators sit at $C$ and at $D$ on this arc. A surveyor measured the angles under which they see the stage, and the angle $\Angle{AOB}$ at the centre:

What do you notice? Would a third spectator on the same arc see the stage under the same angle?

Answer

Both spectators see the stage under the same angle, $40^\circ$, which is half the angle $\Angle{AOB} = 80^\circ$ at the centre. This is no coincidence: every point of the arc sees $\Segment{AB}$ under half the angle at the centre (the theorems “Angle at the Centre” and “Angles on the Same Arc”), so a third spectator on the arc would also see the stage under $40^\circ$. The seats are placed so that everyone has the same view of the stage.

Definition — Angle at the Centre and Angle at the Circumference

Let $A$ and $B$ be two points of a circle of centre $O$, and consider one of the two arcs $AB$. The angle at the centre subtended by this arc is the angle $\Angle{AOB}$ whose inside contains the arc; it is a reflex angle when the arc is a major arc. For a point $C$ of the circle that is not on this arc, the angle $\Angle{ACB}$ is an angle at the circumference (or inscribed angle) subtended by the same arc.

Here the arc $AB$ drawn in bold is subtended by the angle at the centre $\Angle{AOB}$ and by the angle at the circumference $\Angle{ACB}$.

Proposition — Angle at the Centre

The angle at the centre is twice any angle at the circumference subtended by the same arc: if $C$ is a point of the circle of centre $O$ not on the arc $AB$, then $$\Angle{AOB} = 2\Angle{ACB}.$$ This also holds when the arc is a major arc, the angle at the centre being then reflex.

Proof

Let $C$ be a point of the circle not on the arc $AB$. The segments $\Segment{OA}$, $\Segment{OB}$ and $\Segment{OC}$ are radii, so a triangle with two of them as sides is isosceles at $O$. Three cases arise, according to the position of the centre $O$ with respect to the angle $\Angle{ACB}$ (figures 1, 2 and 3).

Case 1: $O$ lies on a side of the angle, say on $\Segment{CB}$, which is then a diameter. The triangle $OCA$ is isosceles at $O$ ($OC = OA$), and, $C$, $O$, $B$ being collinear, $\Angle{AOB}$ is an exterior angle of this triangle: $$\begin{aligned}[t] \Angle{AOB} &= \Angle{OCA} + \Angle{OAC} && \text{(exterior angle of the triangle } OCA\text{)}\\ &= 2\Angle{OCA} && \text{(base angles of the isosceles triangle } OCA\text{)}\\ &= 2\Angle{ACB}. \end{aligned}$$ Case 2: $O$ lies inside the angle $\Angle{ACB}$. Let $\Segment{CD}$ be the diameter through $C$. The ray $\Ray{CD}$ cuts $\Angle{ACB}$ into the angles $\Angle{ACD}$ and $\Angle{DCB}$, each of which has the diameter $\Segment{CD}$ as a side. By case 1, $\Angle{AOD} = 2\Angle{ACD}$ and $\Angle{DOB} = 2\Angle{DCB}$, so $$\begin{aligned}[t] \Angle{AOB} &= \Angle{AOD} + \Angle{DOB}\\ &= 2\Angle{ACD} + 2\Angle{DCB}\\ &= 2\Angle{ACB}. \end{aligned}$$ When $C$ lies on the minor arc, the arc $AB$ not containing $C$ is the major arc, $O$ lies inside the angle $\Angle{ACB}$, and the same computation gives the reflex angle $\Angle{AOB}$. Case 3: $O$ lies outside the angle $\Angle{ACB}$. Let $\Segment{CD}$ be the diameter through $C$, and name the points so that the ray $\Ray{CA}$ lies between the rays $\Ray{CD}$ and $\Ray{CB}$. By case 1, $\Angle{DOB} = 2\Angle{DCB}$ and $\Angle{DOA} = 2\Angle{DCA}$, so $$\begin{aligned}[t] \Angle{AOB} &= \Angle{DOB} - \Angle{DOA}\\ &= 2\Angle{DCB} - 2\Angle{DCA}\\ &= 2\Angle{ACB}. \end{aligned}$$ In every case, the angle at the centre is twice the angle at the circumference.

Method — Finding Angles in a Circle

Mark the centre and the radii: a triangle with two radii as sides is isosceles, and its base angles are equal.
For each angle at the circumference, name the arc it stands on: the angle at the centre on the same arc is twice as large.
Write one fact per line, each followed by its reason: a circle theorem called by its name, a property of the triangle (angle sum, isosceles triangle, exterior angle), or angles on a straight line.
Check: the angles of each triangle of the figure add up to $180^\circ$.

Example

The points $A$, $B$, $C$ and $D$ lie on the circle of centre $O$, with $C$ on the major arc $AB$ and $D$ on the minor arc $AB$, and $\Angle{OAB} = 26^\circ$.

Find the angles $\Angle{AOB}$, $\Angle{ACB}$ and $\Angle{ADB}$, giving a reason for each step.

Answer

$$\begin{aligned}[t] \Angle{OBA} &= \Angle{OAB} = 26^\circ && \text{(}OA = OB\text{: the triangle } OAB \text{ is isosceles)}\\ \Angle{AOB} &= 180^\circ - 26^\circ - 26^\circ && \text{(angle sum of the triangle } OAB\text{)}\\ &= 128^\circ\\ \Angle{ACB} &= \frac{128^\circ}{2} = 64^\circ && \text{(angle at the centre: } C \text{ is on the major arc)}\\ \Angle{ADB} &= \frac{\text{reflex } \Angle{AOB}}{2} = \frac{360^\circ - 128^\circ}{2} && \text{(angle at the centre: } D \text{ is on the minor arc)}\\ &= 116^\circ. \end{aligned}$$ Check: the two angles at the centre fill the full turn, $2 \times 64^\circ + 2 \times 116^\circ = 128^\circ + 232^\circ = 360^\circ$.

Example

The chord $\Segment{AB}$ of a circle of centre $O$ has the same length as the radius.

Prove that every angle at the circumference subtended by the minor arc $AB$ measures $30^\circ$.

Answer

Given: $AB = OA = OB$. To prove: $\Angle{ACB} = 30^\circ$ for every point $C$ of the major arc $AB$. $$\begin{aligned}[t] OA &= OB = AB && \text{(radii, and the datum)}\\ \Angle{AOB} &= 60^\circ && \text{(the triangle } OAB \text{ is equilateral)}\\ \Angle{ACB} &= \frac{60^\circ}{2} = 30^\circ && \text{(angle at the centre)} \end{aligned}$$ Conclusion. Every angle at the circumference subtended by the minor arc $AB$ measures $30^\circ$.

Skills to practice

Finding Angles at the Centre
Proving with the Angle at the Centre

I.2 Angles on the Same Arc

Proposition — Angles on the Same Arc

Angles at the circumference subtended by the same arc are equal: if $C$ and $D$ are two points of a circle that are not on the arc $AB$, then $$\Angle{ACB} = \Angle{ADB}.$$

Proof

Let $O$ be the centre of the circle. The angles $\Angle{ACB}$ and $\Angle{ADB}$ are subtended by the same arc $AB$, so by the theorem “Angle at the Centre” each of them is half of the same angle at the centre $\Angle{AOB}$: $$\Angle{ACB} = \frac{1}{2}\Angle{AOB} = \Angle{ADB}.$$

The Angle in a Semicircle Revisited

When the arc $AB$ is a semicircle, the angle at the centre subtended by it is a straight angle, $180^\circ$: every angle at the circumference subtended by a semicircle measures $\dfrac{180^\circ}{2} = 90^\circ$. The theorem “Angle in a Semicircle” is the special case of the angle at the centre in which the chord is a diameter.

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Example

The points $A$, $B$, $C$ and $D$ lie on a circle, and the chords $\Segment{AC}$ and $\Segment{BD}$ cut at $X$. Moreover $\Angle{ABD} = 32^\circ$ and $\Angle{BDC} = 47^\circ$.

Find the angles $\Angle{ACD}$, $\Angle{BAC}$ and $\Angle{AXB}$, giving a reason for each step.

Answer

$$\begin{aligned}[t] \Angle{ACD} &= \Angle{ABD} = 32^\circ && \text{(angles on the same arc } AD\text{)}\\ \Angle{BAC} &= \Angle{BDC} = 47^\circ && \text{(angles on the same arc } BC\text{)}\\ \Angle{AXB} &= 180^\circ - 47^\circ - 32^\circ && \text{(angle sum of the triangle } ABX\text{)}\\ &= 101^\circ. \end{aligned}$$ Check in the triangle $DXC$: $\Angle{DXC} = \Angle{AXB} = 101^\circ$ (vertically opposite angles), and $32^\circ + 47^\circ + 101^\circ = 180^\circ$.

Example

The chords $\Segment{AC}$ and $\Segment{BD}$ of a circle cut at $X$.

Prove that the triangles $XAB$ and $XDC$ are similar, then that $XA \times XC = XB \times XD$.

Answer

Similar triangles. $$\begin{aligned}[t] \Angle{XAB} = \Angle{CAB} &= \Angle{CDB} = \Angle{XDC} && \text{(angles on the same arc } BC\text{)}\\ \Angle{AXB} &= \Angle{DXC} && \text{(vertically opposite angles)} \end{aligned}$$ The triangles $XAB$ and $XDC$ have two pairs of equal angles: they are similar, with $X$, $A$, $B$ corresponding to $X$, $D$, $C$. The product. In similar triangles, corresponding sides are proportional: $$\begin{aligned}[t] \frac{XA}{XD} &= \frac{XB}{XC} && \text{(similar triangles } XAB \text{ and } XDC\text{)}\\ XA \times XC &= XB \times XD && \text{(cross-multiplying)}. \end{aligned}$$

Skills to practice

Finding Angles on the Same Arc
Proving with Angles on the Same Arc

II Cyclic Quadrilaterals and Tangents

II.1 Cyclic Quadrilaterals

Discover

Four points $A$, $B$, $C$ and $D$ lie, in this order, on a circle. The angles of the quadrilateral $ABCD$ were measured:

Calculate the sums of opposite angles $\Angle{DAB} + \Angle{BCD}$ and $\Angle{ABC} + \Angle{CDA}$. What do you notice? Would it be the same for any four points of a circle, taken in order?

Answer

$\Angle{DAB} + \Angle{BCD} = 66^\circ + 114^\circ = 180^\circ$ and $\Angle{ABC} + \Angle{CDA} = 102^\circ + 78^\circ = 180^\circ$: in each pair, the two opposite angles are supplementary. This is no coincidence: the angles $\Angle{DAB}$ and $\Angle{BCD}$ stand on the two arcs $BD$, whose angles at the centre together make a full turn, $360^\circ$; each of them is half of its angle at the centre, so together they make $180^\circ$ (the theorem “Opposite Angles of a Cyclic Quadrilateral”). The same holds for any four points of a circle taken in order.

Definition — Cyclic Quadrilateral

A quadrilateral is cyclic when its four vertices lie on one circle; the quadrilateral is then said to be inscribed in the circle.

Proposition — Opposite Angles of a Cyclic Quadrilateral

In a cyclic quadrilateral $ABCD$, the opposite angles are supplementary: $$\Angle{DAB} + \Angle{BCD} = 180^\circ \quad\text{and}\quad \Angle{ABC} + \Angle{CDA} = 180^\circ.$$ Consequently, an exterior angle of a cyclic quadrilateral is equal to the interior opposite angle: if $E$ lies on the line $\Line{DC}$ beyond $C$, then $\Angle{BCE} = \Angle{DAB}$.

Proof

Let $O$ be the centre of the circle. The vertices $A$ and $C$ lie on the two different arcs $BD$. The angle $\Angle{DAB}$ is subtended by the arc $BD$ that contains $C$, and $\Angle{BCD}$ by the arc $BD$ that contains $A$; the angles at the centre subtended by these two arcs together make a full turn:

$$\begin{aligned}[t] 2\Angle{DAB} + 2\Angle{BCD} &= 360^\circ && \text{(angle at the centre, twice)}\\ \Angle{DAB} + \Angle{BCD} &= 180^\circ. \end{aligned}$$ The angles of the quadrilateral add up to $360^\circ$, so $\Angle{ABC} + \Angle{CDA} = 360^\circ - 180^\circ = 180^\circ$. Finally, $D$, $C$, $E$ being collinear, $\Angle{BCE} = 180^\circ - \Angle{BCD} = \Angle{DAB}$ (angles on a straight line).

Proposition — Circumscribed circle of a triangle

Let $ABC$ be a triangle. The perpendicular bisectors of its three sides pass through a single point $O$, which is at equal distance from the three vertices: $OA = OB = OC$. The circle with centre $O$ passing through $A$, $B$ and $C$ is called the circumscribed circle of the triangle.

Proposition — A Test for a Cyclic Quadrilateral

If a convex quadrilateral has two opposite angles that are supplementary, then it is cyclic. Equivalently, a convex quadrilateral with an exterior angle equal to the interior opposite angle is cyclic.

Proof

Let $ABCD$ be a convex quadrilateral with $\Angle{ABC} + \Angle{CDA} = 180^\circ$, and $\Gamma$ the circumscribed circle of the triangle $ABC$, which passes through $A$, $B$ and $C$ (circumscribed circle of a triangle). Suppose, for contradiction, that $D$ is not on $\Gamma$. The diagonals of the convex quadrilateral cut at a point $X$ of the chord $\Segment{AC}$, inside $\Gamma$; so the ray $\Ray{BD}$, which passes through $X$, meets $\Gamma$ again at a point $D'$ beyond $X$. The quadrilateral $ABCD'$ is cyclic, so $\Angle{ABC} + \Angle{CD'A} = 180^\circ$ (opposite angles), hence $\Angle{CD'A} = \Angle{CDA}$. The points $D$ and $D'$ are distinct and both lie on the ray $\Ray{XD}$.

If $D$ lies between $X$ and $D'$ ($D$ inside $\Gamma$, as in the figure), $D$ lies inside the triangle $AD'C$, and the line $\Line{AD}$ cuts $\Segment{CD'}$ at a point $E$. The angle $\Angle{ADC}$ is an exterior angle of the triangle $DEC$, so $\Angle{ADC} \gt \Angle{DEC}$; and $\Angle{AEC} = \Angle{DEC}$ is an exterior angle of the triangle $AED'$, so $\Angle{AEC} \gt \Angle{AD'E} = \Angle{AD'C}$. Hence $\Angle{ADC} \gt \Angle{AD'C}$.
If $D'$ lies between $X$ and $D$ ($D$ outside $\Gamma$), $D'$ lies inside the triangle $ADC$, and the line $\Line{AD'}$ cuts $\Segment{CD}$ at a point $E'$. The angle $\Angle{AD'C}$ is an exterior angle of the triangle $D'E'C$, so $\Angle{AD'C} \gt \Angle{D'E'C}$; and $\Angle{AE'C} = \Angle{D'E'C}$ is an exterior angle of the triangle $AE'D$, so $\Angle{AE'C} \gt \Angle{ADE'} = \Angle{ADC}$. Hence $\Angle{AD'C} \gt \Angle{ADC}$.

Both contradict $\Angle{CD'A} = \Angle{CDA}$. Hence the assumption is false: $D$ lies on $\Gamma$, and $ABCD$ is cyclic. The second form of the test follows: an exterior angle equal to the interior opposite angle makes two opposite angles supplementary.

Method — Proving That Four Points Lie on a Circle

To prove that the points $A$, $B$, $C$, $D$ lie on one circle:

name the quadrilateral with its vertices in order around it, and check that it is convex;
find two opposite angles, each with its reason, and show that they add up to $180^\circ$ — or show that an exterior angle equals the interior opposite angle;
conclude with the test for a cyclic quadrilateral.

When two opposite vertices $B$ and $D$ both see the segment $\Segment{AC}$ at a right angle, the test applies directly ($90^\circ + 90^\circ = 180^\circ$): $B$ and $D$ lie on the circle of diameter $\Segment{AC}$.

Example

The quadrilateral $ABCD$ is cyclic, with $\Angle{DAB} = 76^\circ$, $\Angle{ABC} = 3y$ and $\Angle{CDA} = 2y + 20^\circ$. The side $\Segment{BC}$ is extended beyond $C$ to a point $E$.

Find $x = \Angle{BCD}$, then $y$, and $z = \Angle{DCE}$, giving a reason for each step.

Answer

$$\begin{aligned}[t] x &= 180^\circ - 76^\circ = 104^\circ && \text{(opposite angles of a cyclic quadrilateral)}\\ 3y + 2y + 20^\circ &= 180^\circ && \text{(opposite angles } \Angle{ABC} \text{ and } \Angle{CDA}\text{)}\\ 5y &= 160^\circ\\ y &= 32^\circ\\ z &= 180^\circ - x = 76^\circ && \text{(angles on a straight line)} \end{aligned}$$ So $\Angle{ABC} = 96^\circ$ and $\Angle{CDA} = 84^\circ$, and the exterior angle $z = 76^\circ$ equals the interior opposite angle $\Angle{DAB}$. Check: $76^\circ + 96^\circ + 104^\circ + 84^\circ = 360^\circ$.

Example

In the triangle $ABC$, the point $D$ lies on $\Segment{AB}$ and the point $E$ on $\Segment{AC}$, both distinct from the vertices, with $\Angle{ADE} = \Angle{ACB}$.

Prove that the points $B$, $C$, $E$ and $D$ lie on one circle.

Answer

Consider the convex quadrilateral $BCED$, with its vertices in this order.

$A$, $D$, $B$ are collinear, with $D$ between $A$ and $B$: the angle $\Angle{ADE}$ is the exterior angle of $BCED$ at $D$.
The interior angle opposite $D$ is the angle at $C$: $\Angle{BCE} = \Angle{ACB}$, since $E$ lies on $\Segment{AC}$.
By hypothesis $\Angle{ADE} = \Angle{ACB} = \Angle{BCE}$: the exterior angle at $D$ equals the interior opposite angle.

By the test for a cyclic quadrilateral, $BCED$ is cyclic: $B$, $C$, $E$ and $D$ lie on one circle.

Skills to practice

Finding Angles in Cyclic Quadrilaterals
Proving with Cyclic Quadrilaterals

II.2 Angle between a Tangent and a Chord

Proposition — Tangent and Radius

Let $\mathcal{C}$ be the circle of centre $O$ and radius $r$, and $A$ a point of $\mathcal{C}$. Recall that a line is tangent to $\mathcal{C}$ at $A$ when $A$ is its only common point with $\mathcal{C}$.

If a line is tangent to $\mathcal{C}$ at $A$, then it is perpendicular to the radius $\Segment{OA}$.
Conversely, the line through $A$ perpendicular to the radius $\Segment{OA}$ is tangent to $\mathcal{C}$ at $A$.

So there is exactly one tangent line to $\mathcal{C}$ at $A$: the perpendicular to $\Line{OA}$ through $A$.

Proposition — Tangent–Chord Angle

Let $\Segment{AB}$ be a chord of a circle and $T$ a point, other than $A$, of the tangent line at $A$. The angle $\Angle{TAB}$ between the tangent and the chord is equal to any angle at the circumference $\Angle{ACB}$, where $C$ is a point of the circle on the other side of $\Line{AB}$ from $T$ (in the alternate segment): $$\Angle{TAB} = \Angle{ACB}.$$

Proof

Let $O$ be the centre of the circle.

Case where $\Angle{TAB}$ is acute. All the points of the circle lie on the same side of the tangent as $O$ (if a point $Q$ of the circle were on the other side, the segment $\Segment{OQ}$ would cut the tangent at a point $M \neq A$ with $OM \lt OQ = r$, whereas every point of the tangent other than $A$ is outside the circle: tangent and radius), and $\Angle{TAO} = 90^\circ$ (tangent and radius), so the ray $\Ray{AB}$ lies inside the right angle $\Angle{TAO}$. The triangle $OAB$ is isosceles at $O$ ($OA = OB$), so $\Angle{OAB} = \dfrac{180^\circ - \Angle{AOB}}{2}$. Hence $$\begin{aligned}[t] \Angle{TAB} &= \Angle{TAO} - \Angle{OAB}\\ &= 90^\circ - \frac{180^\circ - \Angle{AOB}}{2} && \text{(tangent and radius; isosceles triangle } OAB\text{)}\\ &= \frac{1}{2}\Angle{AOB}. \end{aligned}$$ The angle $\Angle{AOB}$ is the angle at the centre subtended by the arc $AB$ lying inside the angle $\Angle{TAB}$, which does not contain $C$; so $\Angle{ACB} = \dfrac{1}{2}\Angle{AOB}$ (angle at the centre), and $\Angle{TAB} = \Angle{ACB}$. Case where $\Angle{TAB}$ is right. The line $\Line{AB}$ is then the perpendicular to the tangent at $A$, which is the line $\Line{OA}$ (tangent and radius): $\Segment{AB}$ is a diameter, and $\Angle{ACB} = 90^\circ$ (angle in a semicircle). Case where $\Angle{TAB}$ is obtuse. Let $T'$ be a point of the tangent on the other side of $A$, and $D$ a point of the circle on the same side of $\Line{AB}$ as $T$. The angle $\Angle{T'AB} = 180^\circ - \Angle{TAB}$ is acute, so by the first case $\Angle{T'AB} = \Angle{ADB}$. The quadrilateral $ADBC$ is cyclic, so $\Angle{ACB} = 180^\circ - \Angle{ADB}$ (opposite angles), that is, $\Angle{ACB} = 180^\circ - \Angle{T'AB} = \Angle{TAB}$.

Example

The triangle $ABC$ is inscribed in a circle of centre $O$, and $T$ is a point of the tangent at $A$, on the other side of $\Line{AB}$ from $C$. Moreover $\Angle{TAB} = 68^\circ$ and $\Angle{BAC} = 50^\circ$.

Find the angles $\Angle{ACB}$, $\Angle{ABC}$ and $\Angle{AOB}$, giving a reason for each step.

Answer

$$\begin{aligned}[t] \Angle{ACB} &= \Angle{TAB} = 68^\circ && \text{(tangent--chord angle)}\\ \Angle{ABC} &= 180^\circ - 68^\circ - 50^\circ && \text{(angle sum of the triangle } ABC\text{)}\\ &= 62^\circ\\ \Angle{AOB} &= 2 \times 68^\circ = 136^\circ && \text{(angle at the centre)} \end{aligned}$$ Check with the tangent on the other side of $A$: the angle between it and the chord $\Segment{AC}$ is $180^\circ - 68^\circ - 50^\circ = 62^\circ$, equal to $\Angle{ABC}$, as the tangent–chord angle predicts.

Example

The triangle $ABC$ is inscribed in a circle, with $AB = AC$.

Prove that the tangent line to the circle at $A$ is parallel to $\Line{BC}$.

Answer

Let $T$ be a point of the tangent at $A$ on the other side of $\Line{AB}$ from $C$. $$\begin{aligned}[t] \Angle{TAB} &= \Angle{ACB} && \text{(tangent--chord angle)}\\ &= \Angle{ABC} && \text{(}AB = AC\text{: base angles of the isosceles triangle } ABC\text{)} \end{aligned}$$ The lines $\Line{AT}$ and $\Line{BC}$ are cut by the line $\Line{AB}$, and the angles $\Angle{TAB}$ and $\Angle{ABC}$ are alternate angles ($T$ and $C$ lie on either side of $\Line{AB}$). They are equal, so $\Line{AT} \parallel \Line{BC}$: the tangent at $A$ is parallel to $\Line{BC}$.

Skills to practice

Finding Angles between a Tangent and a Chord
Proving with the Tangent-Chord Angle

III Going further

Skills to practice

Solving Problems with Circle Theorems
Investigating Patterns
Solving Problems in Context
Quiz