Most geometry so far has involved triangles and quadrilaterals, which are formed by intervals on lines, and we turn now to the geometry of circles. Lines and circles are the most elementary figures of geometry − a line is the locus of a point moving in a constant direction, and a circle is the locus of a point moving at a constant distance from some fixed point − and all our constructions are done by drawing lines with a straight edge and circles with compasses. Tangents are introduced in this module, and later tangents become the basis of differentiation in calculus.
The theorems of circle geometry are not intuitively obvious to the student, in fact most people are quite surprised by the results when they first see them. They clearly need to be proven carefully, and the cleverness of the methods of proof developed in earlier modules is clearly displayed in this module. The logic becomes more involved − division into cases is often required, and results from different parts of previous geometry modules are often brought together within the one proof. Students traditionally learn a greater respect and appreciation of the methods of mathematics from their study of this imaginative geometric material.
The theoretical importance of circles is reflected in the amazing number and variety of situations in science where circles are used to model physical phenomena. Circles are the first approximation to the orbits of planets and of their moons, to the movement of electrons in an atom, to the motion of a vehicle around a curve in the road, and to the shapes of cyclones and galaxies. Spheres and cylinders are the first approximation of the shape of planets and stars, of the trunks of trees, of an exploding fireball, and of a drop of water, and of manufactured objects such as wires, pipes, ball-bearings, balloons, pies and wheels.
Radii and chords
We begin by recapitulating the definition of a circle and the terminology used for circles. Throughout this module, all geometry is assumed to be within a fixed plane.
Symmetries of a circle
Circles have an abundance of symmetries:
As a result of these symmetries, any point P on a circle
the plane has this property except for lines.
Congruence and similarity of circles
More generally, any two circles are similar − move one circle so that its centre coincides with the centre of the other circle, then apply an appropriate enlargement so that it coincides exactly with the second circle.
A circle forms a curve with a definite length, called the circumference, and it encloses a definite area. The similarity of any two circles is the basis of the definition of π, the ratio of the circumference and the diameter of any circle. We saw in the module, The Circles that if a circle has radius r, then
circumference of the circle = 2πr and area of the circle = πr2
Radii and chords
In the module, Rhombuses, Kites and Trapezia we discussed the axis of symmetry
Let AB be a chord of a circle with centre O. The following three lines coincide:
Trigonometry and chords
Constructions with radii and chords give plenty of opportunity for using trigonometry.
In the diagram to the right, the interval AM subtends an angle θ
at the centre O of a circle of radius 1. Show that:
a AM = sin θ b OM = cos θ
This exercise shows that sine can be regarded as the length of the semichord AM in a circle of radius 1, and cosine as the perpendicular distance of the chord from the centre. Until modern times, tables of sines were compiled as tables of chords or semichords, and the name ‘sine’ is conjectured to have come in a complicated and confused way from the Indian word for semichord.
Arcs and sectors
Now join the radii OA and OB. The reflex angle
The two radii divide the circle into two sectors, called correspondingly the major sector OAB and the minor sector OAB.
It is no surprise that equal chords and equal arcs both subtend equal angles at the centre of a fixed circle. The result for chords can be proven using congruent triangles, but congruent triangles cannot be used for arcs because they are not straight lines, so we need to identify the transformation involved.
Theorema Equal chords of a circle subtend equal angles at the centre.
b Equal arcs of a circle subtend equal angles at the centre.
In the diagram to the right,
POQ (matching angles of congruent
triangles) b Rotate the circle so that the arc PQ coincides with the
arc AB or BA. Then the angles
and hence are equal.
A chord AB of a circle divides the circle into two segments. If AB is a diameter, the two congruent segments are called semicircles − the word ‘semicircle’ is thus used both for the semicircular arc, and for the segment enclosed by the arc and the diameter. Otherwise, the two segments are called a major segment and a minor segment.
The word ‘subtend’
The word ‘subtend’ literally means ‘holds under’, and is often used in geometry to describe an angle.
Suppose that we have an interval or arc AB and a point P not on AB. Join the intervals AP and BP to form the angle
Angles in a semicircle
An angle in a semicircle is a right angle.
Join the radius PO, and let α =
The triangles AOP and BOP are isosceles because all radii are equal, so
Hence α + β + (α + β ) = 180° (angle sum of
This famous theorem is traditionally ascribed to the Greek mathematician Thales, the first known Greek mathematician.
A quadrilateral whose diagonals are equal and bisect each other is a rectangle.
Join the radius PO and produce it to a diameter POQ, then join up the quadrilateral APBQ. Explain why
Constructing a right angle at the endpoint of an interval
1 Choose a point O above the interval AX.
2 Draw a circle with centre O through A crossing
AX again at P.
3 Draw the radius PO and produce it to a diameter POQ.
4 Join AQ, then
PAQ = 90°.
The converse theorem
The angle in a semicircle theorem has a straightforward converse that is best expressed as a property of a right-angled triangle:
We need to prove that MC = MA = MB. Complete the rectangle ACBR. Because ACBR is a rectangle, its diagonals bisect each other and are equal. Hence M is the midpoint of the other diagonal CR, and AM = BM = CM = RM .
This is an excellent example of the way ideas in geometry fit together − a significant theorem about circles has been proven using a property of rectangles.
Two practical situations to illustrate the converse theorem
A set of points in the plane is often called a locus. The term is used particularly when the set of points is the curve traced out by a moving point. For example, a circle can be defined as the locus of a point that moves so that its distance from some fixed point is constant. The two examples below use the converse of the angle in a semicircle theorem to describe a locus.
of all positions where he can stand.
Angles at the centre and circumference
The angle-in-a-semicircle theorem can be generalised considerably. In each diagram below, AB is an arc of a circle with centre O, and P is a point on the opposite arc. The arc AB subtends the angle
In the middle diagram, where the arc is a semicircle, the angle at the centre is a straight angle, and by the previous theorem, the angle at the circumference is a right angle − exactly half. We shall show that this relationship holds also for the other two cases, when the arc is a minor arc (left-hand diagram) or a major arc (right-hand diagram). The proof uses isosceles triangles in a similar way to the proof of Thales’ theorem.
An angle at the circumference of a circle is half the angle at the centre subtended by the same arc.
Let AB be an arc of a circle with centre O, and let P be any point on the opposite arc. We need to prove
Case 1: O lies inside
Case 1: Join PO and produce PO to Q. Then OA = OB = OP (radii), so we have two isosceles triangles OAP and OAQ.
Complete the proof in the other two cases.
A punter stands on the edge of a circular racing track. With his binoculars he is following a horse that is galloping around the track at one revolution a minute. Explain why the punter’s binoculars are rotating at a constant rate of half a revolution per minute.
Angles subtended by the same arc
This corollary of the previous theorem is a particularly significant
Two angles at the circumference subtended by the same arc are equal.
Thales’ theorem is a special case of this theorem.
Some alternative terminology
The last two theorems are often expressed in slightly different language, and some explanation is needed to avoid confusion.1 An angle subtended by an arc is often said to be standing on the arc. With this terminology, the two theorems become: - An angle at the circumference of a circle is half the angle at the centre standing on the same arc. - Two angles at the circumference standing on the same arc are equal.
In the context of these two theorems, it is best to avoid the phrases ‘standing on a chord AB’ and ‘subtended by a chord AB′, because we need to distinguish between angles subtended by the major arc AB and angles subtended by the minor arc AB.2
In the upper diagram to the right, the angle
APB is called
an angle in the (major) segment. Notice that it actually stands on the minor arc AB, which can be confusing. We have already used this terminology before in speaking about an ‘angle in a semicircle’. With this terminology, the last theorem becomes:
Two angles in the same segment of a circle are equal.
The situation is illustrated in the lower diagram to the right.
Extension − The orthocentre of a triangle
An altitude of a triangle is a perpendicular from any of the three vertices to the opposite side, produced if necessary. The two cases are illustrated in the diagrams below.
There are three altitudes in a triangle. The following theorem proves that they concurrent at a point called the orthocentre H of the triangle. It is surprising that circles can be used to prove the concurrence of the altitudes.
a Explain why Q, H, P and C are concyclic, and draw the circle.
b Explain why A, B, P and Q are concyclic, and draw the circle.
c Join the common chord PQ, and explain why
d Explain why
ACR to explain why CR ⊥ AB.
When there are four points, we can always draw a circle through any three of them (provided they are not collinear), but only in special cases will that circle pass through the fourth point. A cyclic quadrilateral is a quadrilateral whose vertices all lie on a circle. This is the last type of special quadrilateral that we shall consider.
Constructing the circumcircle of a cyclic quadrilateral
The opposite angles of a cyclic quadrilateral
The distinctive property of a cyclic quadrilateral is that its opposite angles are supplementary. The following proof uses the theorem that an angle at the circumference is half the angle at the centre standing on the same arc.
The opposite angles of a cyclic quadrilateral are supplementary.
Let ABCD be a cyclic quadrilateral with O the centre of the circle.
Join the radii OB and OD. Let α and g be the angles at the centre, as shown on the diagram.
Here is an alternative proof using the fact that two angles in the same segment are equal.
Join the diagonals AC and BD of the cyclic quadrilateral ABCD. Let α, β, γ and δ be the angles shown.a Give a reason why
DBC = α
b What other angles have sizes β, γ and δ?
c Prove that α+ β + γ +δ = 180°.
d Hence prove that the opposite angles of ABCD
Exterior angles of a cyclic quadrilateral
An exterior angle of a cyclic quadrilateral is supplementary to the adjacent interior angle, so is equal to the opposite interior angle. This gives us the corollary to the cyclic quadrilateral theorem:
In the diagram to the right, BC is produced to P to form the exterior angle
If a cyclic trapezium is not a rectangle, show that the other two sides are not parallel, but have equal length.
Extension − A test for a cyclic quadrilateral
The property of a cyclic quadrilateral proven earlier, that its opposite angles are supplementary, is also a test for a quadrilateral to be cyclic. That is the converse is true. This theorem completes the structure that we have been following − for each special quadrilateral, we establish its distinctive properties, and then establish tests for it.
The proof uses ‘proof by contradiction’, and is thus a little more difficult than other
If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.
Let ABCD be a quadrilateral with
Construct the circle through A, B and D, and suppose, by way of contradiction, that the circle does not pass through C.
Let DC, produced if necessary, meet the circle again at X, and join XB.
so the angles
Hence XB and CB are the same line, so C and X coincide, that is the circle does pass through C.
Prove the following alternative form of the above theorem:
If an exterior angle of a quadrilateral equals the opposite interior angle, then the quadrilateral is cyclic.
Prove that the trapezium is cyclic.
Extension − The sine rule and circle geometry
Each term is the ratio of a length over a pure number, so their common value seems to be a length. Thus it reasonable to ask, what is this common length? The answer is a surprise − the common length is the diameter of the circumcircle through
The proof of this result provides a proof of the sine rule that is independent of the proof given in the module,
In any triangle ABC,
Let O be the centre of the circumcircle through A, B and C, and let
It is sufficient to prove that
There are three cases, as shown below.
In cases 1 and 2, construct the diameter BOP, and join PC .
a Complete the following steps of the proof in Case 1.
i Explain why
iiComplete the proof using
b Complete the following steps of the proof in Case 2.
i Explain why
iiComplete the proof using
c Complete the proof in Case 3.
Tangents to circles
A tangent to a circle is a line that meets the circle at just one point. The diagram below shows that given a line and a circle, can arise three possibilities:
The words ‘secant’ and ‘tangent’ are from Latin − ‘secant’ means ‘cutting’ (as in ‘cross-section’), and ‘tangent’ means ‘touching’ (as in ‘tango’).
The point where a tangent touches a circle is called a point of contact. It is not immediately obvious how to draw a tangent at a particular point on a circle, or even whether there may be more than one tangent at that point. The following theorem makes the situation clear, and uses Pythagoras’ theorem in its proof.
Let T be a point on a circle with centre O.a The line through T, perpendicular to the radius OT, is a tangent to the circle.
b This line is the only tangent to the circle at T.
c Every point on the tangent, except for T itself, lies outside the circle.
First we prove parts a and c. Let
be the line through T perpendicular to the radius OT. Let P be any other point on
, and join the interval OP. Then using Pythagoras’ theorem in
OP2 = OT2 + PT2; which is greater than OT2,
so OP is greater than the radius OT. Hence P lies outside the circle, and not on it. This proves that the line
It remains to prove part b, that there is no other tangent
OT2 = OM2 + MT2 = OM2 + MU2 = OU2,
So OU = OT. Hence U also lies on the circle, contradicting the fact that t is a tangent.
Construction − Tangents from an external point
tangents to a circle with centre O from a point P outside
the circle. 1 Join OP and construct the midpoint M of OP .
2 Draw the circle with centre M through O and P, and let it meet the circle at T and U.
Tangents from an external point have equal length
It is also a simple consequence of the radius-and-tangent theorem that the two tangents PT and PU have equal length.
Notice that here, and elsewhere, we are using the word ‘tangent’ in a second sense, to mean not the whole line, but just the interval from an external point to the point
Tangents to a circle from an external point have equal length.
Prove this result using either congruence or Pythagoras’ theorem.
Tangents and trigonometry
The right angle formed by a radius and tangent gives further opportunities for simple trigonometry. The following exercise shows how the names ‘tangent’ and ‘secant’, and their abbreviations tan θ and sec θ, came to be used in trigonometry. In a circle of radius 1, the length of a tangent subtending an angle θ at the centre is tan θ , and the length of the secant from the external point to the centre is sec θ .
PT = tan θ and PO = sec θ
where the trigonometry function secant is defined
Quadrilaterals with incircles
The following exercise involves quadrilaterals within which an incircle can be drawn tangent to all four sides. These quadrilaterals form yet another class of special quadrilaterals.
Common tangents and touching circles
A line that is tangent to two circles is called a common tangent to the circles. When the points of contact are distinct, there are two cases, as in the diagrams below. The two circles lie on the same side of a direct common tangent, and lie on opposite sides of an indirect common tangent.
Two circles are said to touch at a common point T if there is a common tangent to both circles at the point T. As in the diagram below, the circles touch externally when they are on opposite sides of the common tangent, and touch internally when they are on the same side of the common tangent.
Provided that they are distinct, touching circles have only the one point in common.
Extension - The incentre of a triangle
The angle bisectors of a triangle are concurrent, and the resulting incentre is the centre of the incircle, that is tangent to all three sides.
We have to prove IC bisects
Join IC, and let α =
Drop perpendiculars IL, IM and IN to BC, CA and AB respectively.
Summary - The four centres of a triangle
This completes the development of the four best-known centres of a triangle. The results, and the associated terminology and notation, are summarised here for reference.
The alternate segment theorem
The left-hand diagram below shows two angles
It seems reasonable from these diagrams that in the limiting case, the angle
An angle between a chord and a tangent is equal to any angle in the alternate segment.
There are two equally satisfactory proofs of this theorem.
Let AB be a chord of a circle and let BU be a tangent at B.
Let P be a point on the arc that is not within the arms of
There are three cases, depending on whether:
Case 1: The centre O lies on the chord AB.
Case 2: The centre O lies outside the arms of
Case 3: The centre O lies within the arms of
In each case, we need to prove that
Case 1 is easily settled:
α = 90° (angle in a semicircle) and
For the other two cases, construct the diameter BOD, and join DA.
Find an alternative proof in cases 2 and 3 by constructing the radii AO and BO and using angles at the centre.
The result in the following exercise is surprising. One would not expect parallel lines to emerge so easily in a diagram with two touching circles.
Similarity and circles
The final theorems in this module combine similarity with circle geometry to produce three theorems about intersecting chords, intersecting secants, and the square on a tangent.
When two chords intersect within a circle, the products of the intercepts are equal.
In the triangles APM and QBM:
so AM × BM = PM × QM.
The very last step is particularly interesting. It converts the equality of two ratios of lengths to the equality of two products of lengths. This is a common procedure when working with similarity.
The sine rule and similarity
Many problems involving similarity can be handled using the sine rule. The exercise below gives an alternative proof of the intersecting chord theorem using the sine rule to deal directly with the ratio of two sides of the triangles.
The remaining two theorems of this section also have alternative proofs using the sine rule.
Use the sine rule in the diagram in the above proof to prove that
Secants from an external point
We are now dealing with secants from an external point. When a secant through an external point M meets a circle at two points A and B, the lengths AM and BM are called the intercepts of the secant from the external point, and as before AM × MB = PM × MQ.
Be careful here, the chord AB is not one of the two intercepts. With this definition of intercept, the previous theorem can now be stated as follows. Its proof is word for word almost the same, apart from the reasons:
The product of the intercepts on two secants from an external point are equal.
Let ABM and PQM be two secants from an external point M, as shown.
Join AP and BQ.
Tangent and secant from an external point
Now imagine the secant MPQ in the previous diagrams rotating until it becomes a tangent at a point T on the circle. As the secant rotates, the length of each intercept PM and QM gets closer to the length of the tangent TM to the circle from M, so the product PM × QM gets closer to the square TM2 of the tangent from M. This is a proof using limits.
When we draw a secant and a tangent from M, we have seen that the product AM × BM equals the square TM2 of the tangent. The geometric proof is similar to the previous two proofs, but it does require the alternate segment theorem to establish the similarity.
The product of the intercepts on a secant from an external point equals the square of the tangent from that point.
Let ABM be a secant, and TM a tangent, from an external point M, as shown. We need to prove that AM × BM = TM2. Join AT and BT.
The concurrence of three common chords
intersecting chord theorem.
Show that the three common chords AB, PQ and ST to the three circles in the diagram
Circle geometry is often used as part of the solution to problems in trigonometry and calculus.
converse of the circle theorems
The circle theorems proven in this module all have dramatic and important converse theorems, which are tests for points to lie on a circle. The proofs of these converses, and their applications, are usually regarded as inappropriate for Years 9−10, apart from the converse of the angle in a semicircle theorem, which was developed within the module. They are so closely related to the material in this module, however, that they have been fully developed in an Appendix.
Tangents to circles
Substituting the equation of the line into the equation of the circle gives
and the only solution is thus x = 1. Hence the line and the circle have only the single point of intersection T(1, 1), proving that the line is a tangent to the circle.
Similarly, the dotted line x + y = 1 is a secant, intersecting the circle in two points, and the dotted line x + y = 3 does not intersect the circle at all.
Tangents to parabolas
For example, the line y = 2x − 1 is a tangent to the graph of the parabola of y = x2 at the point P(1, 1), because solving the equations simultaneously gives
Hence P(1, 1) is the only point of intersection.
The following diagram shows that even with
Applications in motion and rates of change
When a stone on a string is whirled around in a circle, then suddenly let go, it flies off at a tangent to the circle (ignoring the subsequent fall to the ground). We can interpret this situation by saying that the tangent describes the direction in which the stone was travelling at the instant when it was released. This leads to the concept of a vector which has both magnitude and direction representing the velocity particle.
The study of motion begins with motion in a straight line, that is, in one dimension. When motion in two dimensions is first considered, circular paths and parabolic paths are the first paths to be considered, because they are reasonably straightforward to describe mathematically, and they arise in so many practical situations.
Vectors and complex numbers
Problems in complex numbers often require locating a set of complex numbers on the complex plane. The following example requires some knowledge of vectors and circle geometry.
Sketch the set of complex numbers such that the ratio
If z = −1, then
So suppose that z is neither 1 nor −1.
Many problems similar to this involve not just the theorems developed in the module, but their converses as well, as developed in the Appendix to this module.
History and applications
Greek geometry was based on the constructions of straight lines and circles, using a straight edge and compasses, which naturally gave circles a central place in their geometry. All the theorems developed in the Content and Appendix of this module
The square of the distance of a point P(x, y) from the origin is x2 + y2 by Pythagoras’ theorem, which means that the equation of the circle with radius a and centre the origin is x2 + y2 = a2.
A circle is a simple closed curve with an inside and an outside, a property that it shares with triangles and quadrilaterals. In three dimensions, spheres, cubes and toruses (doughnuts) have an inside and an outside, but a torus is clearly connected in a different way from a sphere. The subject called topology, begun by Euler and developed extensively in the 20th century, begins with such observations.
Greek astronomy made great use of circles and spheres. They knew that the Earth was round, and were able to calculate its circumference with reasonable accuracy. Ptolemy described the heavenly bodies in terms of concentric spheres on which the Moon, the planets, the Sun and the stars were embedded. When astronomy was reconstructed in the 16th century by Copernicus, the heavenly spheres disappeared, but he still used circles for orbits, with the centre moved from the Earth to the Sun. Ellipses and other refinements of the orbits were soon introduced into this basic model of circles by Kepler, who empirically found three laws of motion for planets. At the end of the 17th century Newton used calculus, his laws of motion and the universal law of gravitation to derive Kepler’s laws.
John Dalton reconstructed chemistry at the start of the 19th century on the basis of atoms, which he regarded as tiny spheres, and in the 20th century, models of circular orbits and spherical shells were originally used to describe the motion of electrons around the spherical nucleus. Thus circles and their geometry have always remained at the heart of theories about the microscopic world of atoms and theories about the cosmos and the universe.
Geometry continues to play a central role in modern mathematics, but its concepts, including many generalisations of circles, have become increasingly abstract. For example, spheres in higher dimensional space came to notice in 1965, when John Leech and John Conway made a spectacular contribution to modern algebra by studying an extremely close packing of spheres in 24-dimensional space.
On the other hand, classical Euclidean geometry in the form presented in this module has nevertheless advanced in modern times − here are three results obtained in recent centuries.
The Euler Line
In 1765, Euler discovered that:
The orthocentre, the circumcentre, and the centroid of any triangle are collinear. The centroid divides the interval joining the circumcentre and the orthocentre in
The line joining these three centres is now called the Euler line. The proof is reasonably straightforward, and is presented in the following exercise.
AGX is similar to
MGO, with similarity ratio 2:1.
b Hence prove that AX || OM.
c Hence explain why AX produced is an altitude of
d Complete the proof is the orthocentre of the triangle.
The nine-point circle
In the early 19th century, Poncelet, Feuerbach and others showed that in any triangle,
That is, these nine points lie on a circle.
He also showed that the centre of this nine-point circle lies on the Euler line, and is the midpoint of the interval joining the circumcentre to the orthocentre.
Again, the proof is straightforward enough to present here as a structured exercise, although proving that the centre of the nine-point circle is the midpoint of OH is rather fiddly.
a Prove that RQ and FG are each parallel
to BC and half its length.
b Prove that RF and QG are each parallel to AH and half its length.
c Hence prove that FGQR is a rectangle.
d Prove similarly that FPQE is a rectangle.
e Let N be the midpoint of FQ, and use the properties of a rectangle to prove that
RN = GN = FN = QN = PN = EN; and hence that R, G, F, Q, P and E are concyclic, lying on a circle with centre N .
f By considering
FQV, prove that N is equidistant from Q and V, and hence that the circle also passes through U, V and W.
The diagram to the right shows the altitude AU, the midpoint P of BC, the midpoint S of UP, and the circumcentre O, orthocentre H, and nine-point centre N.
i Explain why SN is the perpendicular bisector of UP.
ii Explain why SN, produced if necessary, bisects OH.
iii Explain why N lies on OH, and is thus its midpoint.
An enlargement transformation associated with Euler’s line and the nine-point circle
Here is a rather dramatic proof using a single enlargement that establishes the existence of the Euler line and the positions on it of the centroid G, the circumcentre O, the orthocentre H, and the nine-point centre N. The notation is the same as that used in exercises 25 and 26.
In this construction, all that is used about the nine-point circle point is that it is the circle through P, Q and R. The fact that the other six points lie on it would be proven afterwards.
The medians trisect each other, so
AG : GP = BG : GQ = CG : GR = 2 : 1.
Consider the enlargement with centre G and enlargement factor −
This enlargement fixes G, and maps A to P, B to Q and C to R, so the circumcircle of
It also follows that the nine-point circle has half the radius of the circumcircle.
Morley’s trisector theorem
The result has many proofs by similar triangles, and we refer the reader particularly to John Conway’s proof and Bollobas’ version.
The result can also be proven using the compound angle formulae of trigonometry, and is thus reasonably accessible to students in senior calculus courses. See
Appendix − Converses to the Circle Theorems
These proofs are best written as proofs by contradiction. The original theorem is used in the proof of each converse theorem. We have seen this approach when Pythagoras’ theorem was used to prove the converse of Pythagoras’ theorem.
The converse of the angle at the centre theorem
The key theorem of the present module is, ‘An angle at the circumference of a circle is half the angle at the centre subtended by the same arc.’ The most straightforward converse of this is:
so the lines BX and PX coincide, and hence the points X and P coincide. Thus P lies on the circle, contradicting the assumption that the circle does not pass through P.
Testing whether four given points are concyclic
Similarly, any three non-collinear points A, B and C are concyclic. When there are four points A, B, C and D, no three collinear, we can ask whether these four points are concyclic, that is, do they lie on a circle.
The remaining converse theorems all provide tests as to whether four given points are concyclic. In each case, we draw the unique circle through three of them and prove that the fourth point lies on this circle.
We shall assume that the fourth point does not lie on the circle and produce a contradiction.
The converse of the angles on the same arc theorem
If an interval subtends equal angles at two points on the same side of the interval, then the two points and the endpoints of the interval are concyclic.
Let AQ, produced if necessary, meet the circle again at X.
A test for a cyclic quadrilateral
We proved earlier, as extension content, two tests for a cyclic quadrilateral:
The proof by contradiction of the first test is almost identical to the proof of the previous converse theorem. The second test is a simple corollary of the first test.
The intersecting chord test
We proved that the product of the intercepts of intersecting chords are equal. The converse of this gives yet another test for four points to be concyclic. The proof proceeds along exactly the same lines.
Let AB and PQ be intervals intersecting at M, with AM × BM = PM × QM. Then the four endpoints A, B, P and Q are concyclic.
Complete the proof of this theorem.
We leave it the reader to formulate and prove the (true) converses to the remaining two theorems about secants from an external point, and a tangent and a secant from an external point. By similar methods, one can also prove the converse of the theorem
ANSWERS TO EXERCISES
EXERCISE 1a Every translation of any distance along
Let O be any point on
and rotation of 180° about O, both map
onto itself (and are identical in their action
on the points of
Reflection of the plane in the line
b The translation of the plane moving P to Q maps
onto itself, and maps P to Q.
Reflection in the perpendicular bisector of the interval PQ maps
Use simple trigonometry in
All radii of a circle are equal, so OA = OB = OP = OQ, so the quadrilateral APBQ is a rectangle because its diagonals are equal and bisect each other. Hence
At all times, the triangle formed by the plank, the wall and the floor is right-angled with hypotenuse of length
Case 2: It will be sufficient to prove the result when O lies on AP.
Case 3: Join PO and produce PO to Q.
EXERCISE 7a The interval HC subtends right angles at P and Q, so the circle with diameter HC passes through them.
b The interval AB subtends right angles at P and Q, so the circle with diameter AB passes through them.
c They stand on the same arc HQ of the first circle.
d They stand on the same arc AQ of the second circle.
e In the triangles ABQ and ACR,
RCA by parts c and d, and
A is common, so
AQB = 90°.
EXERCISE 8a The angles
DBC stand on the same arc DC, so
DBC = α.
BDC = β (angles on the same arc BC),
ABD = γ (angles on the same arc AD),
ADB = δ (angles on the same arc AB).
c 2α + 2β + 2γ + 2δ= 360° (angle sum of quadrilateral ABCD)
The rest is clear.
Join the common chord BQ, and produce ABC to X, and let a =
The adjacent interior angle is supplementary to the exterior angle, and therefore equal to the opposite interior angle. Hence the quadrilateral is cyclic by the theorem.
Hence the quadrilateral ABCD is cyclic (opposite angles are supplementary).
Let tangents from an external point P touch the circle at T and U.
Join the radii OT and OU and the interval OP.
Since OP is common, the radii are equal, and the radii are perpendicular to the tangents,
The triangle OPT is right-angled at T. The rest is simple trigonometry.
EXERCISE 17a i 0 ii 1 iii2 iv3 v4.
b The first diagram below illustrates the situation with two indirect common tangents.
The indirect tangents AT and BU intersect at M. Then AM = BM and TM = UM because tangents from an external point are equal, and adding the lengths gives AT = BU.
c With direct common tangents, there are two cases to consider. Suppose first that the direct tangents AT and BU intersect at M when produced, as in the second diagram above. Then again AM = BM and TM = UM, and subtracting gives AT = BU.
Now suppose that the direct common tangents AT and BU are parallel, as in the third diagram. The radii AO and BO are perpendicular to the parallel tangents, so they lie on the one line and form a diameter, and the radii TZ and UZ also form a diameter. Hence ABUT is a rectangle, so its opposite sides AT and BU are equal.
Use the AAS test.
b They are matching sides of congruent triangles.
c Use the RHS test.
MCI, so IC is also an angle bisector, thus proving the concurrence of the angle bisectors. Also the circle with centre I and radius LI = MI = NI is tangent to all three sides by the radius-and-tangent theorem.
We have already proven that
The sine rule
Using this form of the sine rule in the two triangles,
Proof: In the triangles APM and QBM :
Proof: In the triangles AQM and PBM :
Proof: In the triangles ATM and TBM:
Let the chords AB and PQ meet at M. Join SM and produce it to meet the circles at X and Y. We need to prove that the points X and Y coincide.
Using the intersecting chord theorem in each circle in turn,
EXERCISE 26a First, AG : GM = 2 : 1 because the centroid divides each median in the ratio 2 : 1.
Secondly, XG : GO = 2 : 1 by the construction of the point X .
AGX is similar to
MGO by the SAS similarity test.
Hence the matching angles
and since these are alternate angles, AX || OM.
c Thus AX meets BC at right angles so X lies on the altitude from A. AX produced is an altitude of
d We have shown X is on one altitude and the same argument shows X is on all three altitudes. Thus X is H orthocentre of the triangle.
EXERCISE 27a &b These follow because the interval joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
c Hence FGQR is a parallelogram, and is a rectangle because AU ⊥ BC
d This follows by a similar argument to parts a − c.
e This follows because the diagonals of each rectangles are equal and bisect each other. Note that FQ is a diagonal of both rectangles.
f Since N is the midpoint of the hypotenuse of the right-angled triangle FQV, it follows by the converse of the angle-in-a-semicircle theorem, proven earlier in this module, that N is equidistant from F, Q and V. Similarly, N is equidistant from E, P and U, and from G, R and W. Hence the circle passes through all nine points.
g i The perpendicular bisector of UP is the locus of all points equidistant from U and P, and this includes the nine-point centre N.
ii Complete the rectangle UPOT. Then NS || HTU, so NS bisects OT using opposite sides of rectangles, and so also bisects OH using a similarity theorem on triangles. Hence NS meets OH at Y .
iii By similar arguments to parts i and ii, the perpendicular bisectors of UP, VQ and WR all pass through N, and all bisect OH. Hence the three perpendicular bisectors are concurrent, and the midpoint of OH and the point N coincide at the intersection of these three lines.
Let PQ, produced if necessary, meet the circle again at X .
Hence X coincides with Q, which is a contradiction.
The Improving Mathematics Education in Schools (TIMES) Project 2009-2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations.
The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations.
© The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICE-EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.