5.2: Euclidean geometry- a brief summary

Philosophy is written in this grand book - I mean the universe - which stands continually open to our gaze, but it cannot be understood unless one first learns to comprehend the language and to interpret the characters in which it is written. It is written in the language of mathematics, and its characters are triangles, circles, and other geometrical figures, without which it is humanly impossible to understand a single word of it; without these, one is wandering about in a dark labyrinth.

Galileo Galilei (1564-1642)

This section provides a detailed, but compressed, outline of an initial formalisation of school geometry - of a kind that one would like good students and all teachers to appreciate. It is unashamedly a semi-formal approach for beginners, not a strictly formal treatment (such as that provided by David Hilbert (1862-1943) in his 1899 book Foundations of Geometry, or in the more detailed exposition by Edwin Moise (1918-1998) Elementary Geometry from an Advanced Standpoint, published in 1963). In particular:

• we work with relatively informal notions of points, lines, and angles in the plane;
• we focus attention on certain simple issues which really matter at school level (such as how points, lines, line segments, and angles are referred to; the notion of a triangle as an ordered triple of vertices; the fact that the vertices of a quadrilateral must be labelled cyclically; etc.);
• we limit the formal deductive structure to just three central criteria, namely the criteria for congruence, for parallels, and for similarity, and show how they allow one to develop results and methods in a logical sequence.

We begin with the intuitive idea of points and lines in the plane. Two points A, B determine

• the line segment AB (with endpoints A and B), and
• the line AB (which extends the line segment AB in both directions - beyond A, and beyond B).

Three points A, B, C determine an angle ∠ABC (between the two line segments BA and BC).

We can then begin to build more complicated figures, such as

• a triangle ABC (with three vertices A, B, C; three sides AB, BC, CA; and three angles ∠ABC at the vertex B, ∠BCA at C, and ∠CAB at A),
• a quadrilateral ABCD (with four vertices A, B, C, D; and four sides AB, BC, CD, DA which meet only at their endpoints).

And so on. Two given points A, B also allow us to construct the circle with centre A, and passing through B (that is, with radius AB).

This very limited beginning already opens up the world of ruler and compasses constructions. In particular, given a line segment AB one can draw:

• the circle with centre A, and passing through B, and
• the circle with centre B, and passing through A.

If the two circles meet at C,

• then AB = AC (radii of the first circle), and BA = BC (radii of the second circle).

Hence we have constructed the equilateral triangle ΔABC on the given segment AB. This construction is the very first proposition in Book 1 of the Elements of Euclid (flourished c. 300 BC). Euclid’s second proposition is presented next as a problem.

Problem 137 Given three points A, B, C, show how to construct - without measuring - a point D such that the segments AB and CD are equal (in length).

Problem 137 looks like a simple starter (where the only available construction is to produce the third vertex of an equilateral triangle on a given line segment). However, to produce a valid solution requires a clear head and a degree of ingenuity.

Given two points A, B, the process of constructing an equilateral triangle ΔABC illustrates how we are allowed to construct new points from old.

• Whenever we construct two lines or circles that cross, the points where they cross (such as the point C in the above construction of the equilateral triangle ΔABC) become available for further constructions. So, if points A and B are given, then once C has been constructed, we may proceed to draw the lines AC and BC.

However, the fact that we can construct a line segment AB does not allow us to ‘measure’ the segment with a ruler, and then to use the resulting measurement to ‘copy’ the segment AB to the point C in order to construct the required point D such that AB = CD. The “ruler” in ruler and compasses constructions is used only to draw the line through two known points - not to measure. (Measuring is an approximate physical action, rather than an exact “mental construction”, and so is not really part of mathematics.) Hence in Problem 137 we have to find another way to produce a copy CD of the segment AB starting at the point C. Similarly, we can construct the circle with centre A and passing through B, but this does not allow us to use the pair of compasses to transfer distances physically (e.g. by picking up the compasses from AB and placing the compass point at C, like using the old geometrical drawing instrument that was called a pair of dividers). In seeking the construction required in Problem 137, we are restricted to “exact mental constructions” which may be described in terms of:

• drawing (or constructing) the line joining any two known points,
• constructing the circle with centre at a known point and passing through a known point, and
• obtaining a new point D as the intersection of two constructed lines or circles (or of a line and a circle).

If on the line AB, the point X lies between A and B, then we obtain a straight angle ∠AXB at X (or rather two straight angles at X - one on each side of the line AB). If we assume that all straight angles are equal, then it follows easily that “vertically opposite angles are always equal”.

Problem 138 Two lines AB and CD cross at X, where X lies between A and B and between C and D. Prove that ∠AXC=∠BXD.

Define a right angle to be ‘half a straight angle’. Then we say that two lines which cross at a point X are perpendicular if an angle at X is a right angle (or equivalently, if all four angles at X are equal). The next step requires us to notice two things - partly motivated by experience when coordinating hand, eye and brain to construct, and to think about, physical structures.

• First we need to recognise that triangles hold the key to the analysis of more complicated shapes.
• Then we need to realise that triangles in different positions can still be “equal”, or congruent - which then focuses attention on the minimal conditions under which two triangles can be guaranteed to be congruent.

The first of these two bullet points has an important consequence - namely that solving any problem in 2- or in 3-dimensions generally reduces to working with triangles. In particular, solving problems in 3-dimensions reduces to working in some 2-dimensional cross-section of the given figure (since three points not only determine a triangle, but also determine the plane in which that triangle lies). It follows that 2-dimensional geometry holds the key to solving problems in 3-dimensions, and that working with triangles is central in all geometry.

The second bullet point forces us to think carefully about:

• what we mean by a triangle (and in particular, to understand why ΔABC and ΔBCA are in some sense different triangles, even though they use the same three vertices and sides), and
• what it means for two triangles to be “the same”.

A triangle ΔABC incorporates six pieces of data, or information: the three sides AB, BC, CA and the three angles ∠ABC, ∠BCA, ∠CAB. We say that two (ordered) triangles ΔABC and ΔAʹBʹCʹ are congruent (which we write as

$$\Delta ABC\equiv \Delta A^{\prime }{B}^{\prime }{C}^{\prime },$$

where the order in which the vertices are listed matters) if their sides and angles “match up” in pairs, so that

$$\begin{array}{l}\underset{\_}{AB}=\underset{\_}{A^{\prime }{B}^{\prime }},\underset{\_}{BC}={\underset{\_}{B^{\prime }{C}^{\prime }}}^{\prime },\underset{\_}{CA}=\underset{\_}{C^{\prime }{A}^{\prime }},\\ \angle ABC=\angle A^{\prime }{B}^{\prime }{C}^{\prime },\angle BCA=\angle B^{\prime }{C}^{\prime }{A}^{\prime },\angle CAB=\angle C^{\prime }{A}^{\prime }{B}^{\prime }.\end{array}$$

As a result of drawing and experimenting with our hands, our minds may realise that certain subsets of these six conditions suffice to imply the others. In particular:

SAS-congruence criterion: if

$$\underset{\_}{AB}=\underset{\_}{A^{\prime }{B}^{\prime }},\angle ABC=\angle A^{\prime }{B}^{\prime }{C}^{\prime },\underset{\_}{BC}=\underset{\_}{B^{\prime }{C}^{\prime }},$$

then

$$\Delta ABC\equiv \Delta A^{\prime }{B}^{\prime }{C}^{\prime }$$

(where the name “SAS” indicates that the three listed match-ups occur in the specified order S (side), A (angle), S (side) as one goes round each triangle).

SSS-congruence criterion: if

$$\underset{\_}{AB}=\underset{\_}{A^{\prime }{B}^{\prime }},\underset{\_}{BC}=\underset{\_}{B^{\prime }{C}^{\prime }},\underset{\_}{CA}=\underset{\_}{C^{\prime }{A}^{\prime }},$$

then

$$\Delta ABC\equiv \Delta A^{\prime }{B}^{\prime }{C}^{\prime }.$$

ASA-congruence criterion: if

$$\angle ABC=\angle A^{\prime }{B}^{\prime }{C}^{\prime },\underset{\_}{BC}=\underset{\_}{B^{\prime }{C}^{\prime }},\angle BCA=\angle B^{\prime }{C}^{\prime }{A}^{\prime },$$

then

$$\Delta ABC\equiv \Delta A^{\prime }{B}^{\prime }{C}^{\prime }.$$

If in a given triangle ΔABC we have AB = AC, then we say that ΔABC is isosceles with apex A, and base BC (iso = same, or equal; sceles = legs).

Problem 139 Let ΔABC be an isosceles triangle with apex A. Let M be the midpoint of the base BC. Prove that ΔAMB = ΔAMC and conclude that AM is perpendicular to the base BC.

Problem 140 Construct two non-congruent triangles, ΔABC and ΔAʹBʹCʹ, where $\angle BCA=\angle B^{\prime }{C}^{\prime }{A}^{\prime }={30}^{°},\left|\underset{\_}{CA}\right|=\left|\underset{\_}{C^{\prime }}\right|=\sqrt{3},\left|\underset{\_}{AB}\right|=\left|\underset{\_}{A^{\prime }{B}^{\prime }}\right|=1.$

Conclude that there is in general no “ASS-congruence criterion”.

The congruence criteria allow one to prove basic results such as:

Claim In any isosceles triangle ΔABC with apex A (i.e. with AB = AC), the two base angles ∠B and ∠C are equal.

Proof 1 Let M be the midpoint of BC.

Then ΔAMB = ΔAMC (by the SSS-congruence criterion, since

AM = AM,

MB = MC (by construction of M as the midpoint)

BA = CA (given)).

$\therefore \angle B=\angle ABM=\angle ACM=\angle C.$ QED

Proof 2 ΔBAC = ΔCAB (by the SAS-congruence criterion, since BA

BA = CA (given),

∠BAC = ∠CAB (same angle),

AC = AB (given),

$\therefore \angle B=\angle ABC=\angle ACB=\angle C.$ QED

We also have the converse result:

Claim In any triangle ΔABC, if the base angles ∠B and ∠C are equal, then the triangle is isosceles with apex A (i.e. AB = AC).

Proof ΔABC = ΔACB (by the ASA-congruence criterion, since

∠ABC = ∠ACB (given),

BC = CB, and

∠BAC = ∠CAB (given)).

$\therefore \underset{\_}{AB}=\underset{\_}{AC}.$ QED

Problem 141

(i) A circle with centre O passes through the point A. The line AO meets the circle again at B. If C is a third point on the circle, prove that ∠ACB is equal to ∠CAB + ∠CAB.

(ii) Conclude that, if the angles in ΔABC add to a straight angle, then ∠ACB is a right angle.

Once we introduce the parallel criterion, and hence can prove that the three angles in any triangle add to a straight angle, Problem 141 will guarantee that “the angle subtended on the circumference by a diameter is always a right angle”.

Problem 142 Show how to implement the basic ruler and compasses constructions:

(i) to construct the midpoint M of a given line segment AB;

(ii) to bisect a given angle ∠ABC;

(iii) to drop a perpendicular from P to a line AB (that is, to locate X on the line AB, so that the two angles that PX makes with the line AB on either side of PX are equal).

Prove that your constructions do what you claim.

Problem 143 Given two points A and B.

(a) Prove that each point X on the perpendicular bisector of AB is equidistant from A and from B (that is, that XA = XB).

(b) Prove that, if X is equidistant from A and from B, then X lies on the perpendicular bisector of AB.

Problem 143 shows that, given a line segment AB, the perpendicular bisector of AB is the locus of all points X which are equidistant from A and from B. This observation is what lay behind the construction of the circumcentre of a triangle (back in Chapter 1, Problem 32(a)):

Given any ΔABC.

Let O be the point where the perpendicular bisectors of AB and BC meet.

Then OA = OB

and OB = OC.

$\therefore \underset{\_}{OA}=\underset{\_}{OB}=\underset{\_}{OC}.$

Hence O is the centre of a circle passing through all three vertices A, B, C.

Moreover O also lies on the perpendicular bisector of CA.

This circle is called the circumcircle of ΔABC, and O is called the circumcentre of ΔABC. As indicated back in Problem 32, the radius of the circumcircle of ΔABC (called the circumradius of the triangle) is generally denoted by R. Later we will meet other circles and “centres” associated with a given triangle ΔABC.

Before moving on it is worth extending Problem 143 to three dimensions

Problem 144 Given any two points N, S in 3D space, prove that the locus of all points X which are equidistant from N and from S form the plane perpendicular to the line NS and passing through the midpoint M of NS.

The next two fundamental results are often neglected.

Problem 145 Given any ΔABC, if we extend the side BC beyond C to a point X, then the “exterior angle” ∠ACX at C is greater than each of the “two interior opposite angles” ∠A and ∠B.

Problem 146

(a) If in ΔABC we have AB > AC, then ∠ACB > ∠ABC. (“In any triangle, the larger angle lies opposite the longer side.”)

(b) If in ΔABC we have ∠ACB > ∠ABC, then AB > AC. (“In any triangle, the longer side lies opposite the larger angle.”)

(c) (The triangle inequality) Prove that in any triangle ΔABC,

$$\underset{\_}{AB}+\underset{\_}{BC}>\underset{\_}{AC}.$$

The results in Problems 145 and 146 have surprisingly many consequences. For example, they allow one to prove the converse of the result in Problem 141

Problem 147 Suppose that in ΔABC, ∠C = ∠A + ∠B. Prove that C lies on the circle with diameter AB.

(In particular, if the angles of ΔABC add to a straight angle, and ∠ACB is a right angle, then C lies on the circle with diameter AB.

We come next to a result whose justification is often fudged. At first sight it is unclear how to begin: there seems to be so little information to work with - just two points and a line through one of the points.

Problem 148 A circle with centre O passes through the point P. Prove that the tangent to the circle at P is perpendicular to the radius OP.

Problem 148 is an example of a result which implies its own converse - though in a backhanded way. Suppose a circle with centre O passes through the point P. If OP is perpendicular to a line m passing through P, then m must be tangent to the circle (because we know that the tangent at P is perpendicular to OP, so the angle between m and the tangent is “zero”, which forces m to be equal to the tangent). This converse will be needed later, when we meet the incircle.

Problem 149 Let P be a point and m a line not passing through P. Prove that, among all possible line segments PX with X on the line m, a perpendicular from P to the line m is the shortest.

The result in Problem 149 allows us to define the “distance” from P to the line m to be the length of any perpendicular from P to m. (As far as we know at this stage of the development, there could be more than one perpendicular from P to m.)

Note that all the results mentioned so far have avoided using the Euclidean “parallel criterion” (or - equivalently - the fact that the three angles in any triangle add to a straight angle). So results proved up to this point should still be “true” in any geometry where we have points, lines, triangles, and circles satisfying the congruence criteria - whether or not the geometry satisfies the Euclidean “parallel criterion”.

The idea that there is only one “shortest” distance from a point to a line may seem “obvious”; but it is patently false on the sphere, where every line (i.e. ‘great circle’) from the North pole P to the equator is perpendicular to the equator (and all these lines have the same “length”). The proof that there is just one such perpendicular from P to m depends on the parallel criterion (see below) - a criterion which fails to hold for geometry on the sphere.

Euclid’s Elements started with a few basic axioms that formalised the idea of ruler and compasses constructions. He then added a simple axiom that allowed one to compare angles in different locations. He made the forgivable mistake of omitting an axiom for congruence of triangles - imagining that it can be proved. (It can’t.) However he then stated, and carefully developed the consequences of, a much more subtle axiom about parallel lines (two lines m, n in the plane are said to be parallel if they never meet, no matter how far they are extended). For reasons that remain unclear, instead of appreciating that Euclid's “parallel postulate” constituted a profound insight into the foundations of geometry, mathematicians in later ages saw the complexity of Euclid’s postulate as some kind of flaw, and so tried to show that it could be derived from the other, simpler postulates. The attempt to “correct” this perceived flaw became a kind of Holy Grail.

The story is instructive, but too complicated to summarise accurately here. The situation was eventually clarified by two nineteenth century mathematicians (more-or-less at the same time, but working independently). In the revolutionary, romantic spirit of the nineteenth century, János Bolyai (Hungarian: 1802-1860) and Nikolai Lobachevski (Russian: 1792-1856) each allowed himself to consider what would happen if one adopted a different assumption about how “parallel lines” behave. Both discovered that one can then derive an apparently coherent theory of a completely novel kind, with its own beautiful results: that is, a geometry which seemed to be internally “consistent” - but different from Euclidean geometry. Lobachevski published brief notes of his work in 1829-30 (in Kazan); Bolyai knew nothing of this and published incomplete notes of his researches in 1832. Lobachevski published a more detailed booklet in 1840.

Neither mathematician got the recognition he might have anticipated, and it was only much later (largely after their deaths) that others realised how to show that the fantasy world they had each dreamt up was just as “internally consistent” as traditional Euclidean geometry. The story is further complicated by the fact that the dominant mathematician of the time - namely Gauss (1777-1855) - claimed to have proved something similar (and he may well have done so, but exactly what he knew has to be inferred from cryptic remarks in occasional letters, since he published nothing on the subject). If there is a moral to the story, it could be that success in mathematics may not be recognised, or may only be recognised after one’s death: so those who spend their lives exploring the mathematical universe had better appreciate the delights of the mathematical journey, rather than being primarily motivated by a desire for immediate recognition and acclaim!

Two lines m, n in the plane are said to be parallel if they never meet - no matter how far they are extended. We sometimes write this as “m||n”.

Given two lines m, n in the plane, a third line p which crosses both m and n is called a transversal of m and n.

Parallel criterion: Given two lines m and n, if some transversal p is such that the two “internal” angles on one side of the line p (that is the two angles that p makes with m and with n, and which lie between the two lines m and n) add to less than a straight angle, then the lines m and n must meet on that side of the line p.

If the internal angles on one side of p add to more than a straight angle, then internal angles on the other side of p add to less than a straight angle, so the lines m and n must meet on the other side of p. It follows.

• that two lines m and n are parallel precisely when the two internal angles on one side of a transversal add to exactly a straight angle.

Parallel lines can be thought of as “all having the same direction”; so it is convenient to insist that “every line is parallel to itself” (even though it has lots of points in common with itself). It then follows

• that, given three lines k, m, n, if k is parallel to m and m is parallel to n, then k is parallel to n; and
• that given a line m and a point P, there is a unique line n through P which is parallel to m.

All this then allows one

• to conclude that, if m and n are any two lines, and p is a transversal, then m and n are parallel if and only if alternate angles are equal (or equivalently, if and only if corresponding angles are equal); and
• to extend the basic ruler and compasses constructions to include the construction:

  “given a line AB and a point P, construct the line through P which is parallel to AB”

  (namely, by first constructing the line PX through P, perpendicular to AB, and then the line through P, perpendicular to PX).

One can then prove the standard result about the angles in any triangle.

Claim The angles in any triangle ΔABC add to a straight angle.

Proof Construct the line m through A that is parallel to BC. Then AB and AC are transversals, which cross both the line m and the line BC, and which make three angles at the point A on m:

• one being just the angle ∠A in the triangle ΔABC,
• one being equal to ∠B (alternate angles relative to the transversal AB) and
• one being equal to ∠C (alternate angles relative to the transversal AC).

The three angles at A clearly add to a straight angle, so the three angles ∠A, ∠B, ∠C also add to a straight angle. QED

Once we know that the angles in any triangle add to a straight angle, we can prove all sorts of other useful facts. One is a simple reformulation of the above Claim.

Problem 150 Given any triangle ΔABC, extend BC beyond C to a point X. Then the exterior angle

$$\angle XCA=\angle A+\angle B.$$

(“In any triangle, each exterior angle is equal to the sum of the two interior opposite angles.”)

Another important consequence is the result which underpins the sequence of “circle theorems”.

Problem 151 Let O be the circumcentre of ΔABC. Prove that

$$\angle AOB=2·\angle ACB.$$

Problem 151 implies that

“the angles subtended by any chord AB on a given arc of the circle are all equal”,

and are equal to exactly one half of the angle subtended by AB at the centre O of the circle. This leads naturally to the familiar property of cyclic quadrilaterals.

Problem 152 Let ABCD be a quadrilateral inscribed in a circle (such a quadrilateral is said to be cyclic, and the four vertices are said to be concyclic - that is, they lie together on the same circle). Prove that opposite angles (e.g. ∠B and ∠D) must add to a straight angle. (Two angles which add to a straight angle are said to be supplementary.)

These results have lots of lovely consequences: we shall see one especially striking example in Problem 164. Meantime we round up our summary of the "circle theorems".

Problem 153 Suppose that the line XAY is tangent to the circumcircle of ΔABC at the point A, and that X and C lie on opposite sides of the line AB. Prove that ∠XAB = ∠ACB.

Problem 154

(a) Suppose C, D lie on the same side of the line AB.

(i) If D lies inside the circumcircle of ΔABC, then ∠ADB > ∠ACB.

(ii) If D lies outside the circumcircle of ΔABC, then ∠ADB < ∠ACB.

(b) Suppose C, D lie on the same side of the line AB, and that ∠ACB = ∠ADB. Then D lies on the circumcircle of ΔABC.

(c) Suppose that ABCD is a quadrilateral, in which angles ∠B and ∠D are supplementary. Then ABCD is a cyclic quadrilateral.

Another result which follows now that we know that the angles of a triangle add to a straight angle is a useful additional congruence criterion - namely the RHS-congruence criterion. This is a ‘limiting case’ of the failed ASS-congruence criterion (see the example in Problem 140). In the failed ASS criterion the given data correspond to two different triangles - one in which the angle opposite the first specified side (the first “S” in “ASS”) is acute, and one in which the angle opposite the first specified side is obtuse. In the RHS-congruence criterion, the angle opposite the first specified side is a right angle, and the two possible triangles are in fact congruent.

RHS-congruence criterion: If ∠ABC and ∠AʹBʹCʹ are both right angles, and BC = BʹCʹ, CA = CʹAʹ, then

$$\Delta ABC\equiv \Delta A^{\prime }{B}^{\prime }{C}^{\prime }.$$

Proof Suppose that AB = AʹBʹ. Then

AB = AʹBʹ

∠ABC = ∠AʹBʹCʹ,

BC = BʹCʹ

Hence we may apply the SAS-congruence criterion to conclude that ΔABC = ΔAʹBʹCʹ.

If on the other hand $AB\ne \underset{\_}{A^{\prime }{B}^{\prime }}$ , we may suppose that BA > BʹAʹ. Now construct Aʹʹ on BA such that BAʹʹ = BʹAʹ. Then

AʹʹB = AʹBʹ,

∠AʹʹBC = ∠AʹBʹCʹ,

BC = BʹCʹ,

$\therefore \Delta A^{\prime \prime }BC\equiv \Delta A^{\prime }{B}^{\prime }{C}^{\prime }$ (by SAS-congruence).

Hence AʹʹC = AʹCʹ = AC, so ΔCAAʹʹ is isosceles.

$$\therefore \angle C{A}^{\prime \prime }A=\angle CA{A}^{\prime \prime }.$$

However, ∠CAʹʹA > ∠CBA (since the exterior angle ∠CAʹʹA in ΔCBAʹʹ must be greater than the interior opposite angle ∠CBA, by Problem 145).

But then the two base angles in the isosceles triangle ΔCAAʹʹ are each greater than a right angle - so the angle sum of ΔCAAʹʹ is greater than a straight angle, which is impossible. Hence this case cannot occur. QED

RHS-congruence seems to be needed to prove the basic result (Problem 161 below) about the area of parallelograms, and this is then needed in the proof of Pythagoras’ Theorem (Problem 18). In one sense RHS-congruence looks like a special case of SSS-congruence (as soon as two pairs of sides in two right angled triangles are equal, Pythagoras’ Theorem guarantees that the third pair of sides are also equal). However this observation cannot be used to justify RHS-congruence if RHS-congruence is needed to justify Pythagoras’ Theorem.

Problem 155 Given a circle with centre O, let Q be a point outside the circle, and let QP, QP' be the two tangents from Q, touching the circle at P and at P'. Prove that QP = QPʹ, and that the line OQ bisects the angle ∠PQPʹ.

Problem 156 You are given two lines m and n crossing at the point B.

(a) If A lies on m and C lies on n, prove that each point X on the bisector of angle ∠ABC is equidistant from m and from n.

(b) If X is equidistant from m and from n, prove that X must lie on one of the bisectors of the two angles at B.

Problem 156 shows that, given two lines m and n that cross at B, the bisectors of the two pairs of vertically opposite angles formed at B form the locus of all points X which are equidistant from the two lines m and n. This allows us to mimic the comments following Problem 143 and so to construct the incentre of a triangle.

Given any ΔABC, let I be the point where the angle bisectors of ∠ABC and ∠BCA meet.

Let the perpendiculars from I to the three sides AB, BC, CA meet the sides at P, Q, R respectively. Then

IP = IQ (since I lies on the bisector of ∠ABC) and

IQ = IR (since I lies on the bisector or ∠BCA).

Hence the circle which has centre I and which passes through P also passes through Q and R.

Moreover, I also lies on the bisector of ∠CAB; and since the radii IP, IQ, IR are perpendicular to the sides of the triangle, the circle is tangent to the three sides of the triangle (by the comments following Problem 148)

This circle is called the incircle of ΔABC, and I is called the incentre of ΔABC. The radius of the incircle of ΔABC is called the inradius, and is generally denoted by r.

A quadrilateral ABCD in which AB ∥ DC and BC ∥ AD is called a parallelogram. A parallelogram ABCD with a right angle is a rectangle. A parallelogram ABCD with AB = AD is called a rhombus. A rectangle which is also a rhombus is called a square.

Problem 157 Let ABCD be a parallelogram.

(i) Prove that ΔABC = ΔCDA, so that each triangle has area exactly half of area(ABCD).

(ii) Conclude that opposite sides of ABCD are equal in pairs and that opposite angles are equal in pairs.

(iii) Let AC and BD meet at X. Prove that X is the midpoint of both AC and BD.

Problem 158 Let ABCD be a parallelogram with centre X (where the two main diagonals AC and BD meet), and let m be any straight line passing through the centre. Prove that m divides the parallelogram into two parts of equal area.

We defined a parallelogram to be “a quadrilateral ABCD in which AB ∥ DC and BC ∥ AD however, in practice, we need to be able to recognise a parallelogram even if it is not presented in this form. The next result hints at the variety of other conditions which allow us to recognise a given quadrilateral as being a parallelogram “in mild disguise”.

Problem 159

(a) Let ABCD be a quadrilateral in which AB ∥ DC, and AB = DC. Prove that BC ∥ AD, and hence that ABCD is a parallelogram.

(b) Let ABCD be a quadrilateral in which AB = DC and BC = AD. Prove that AB ∥ DC, and hence that ABCD is a parallelogram.

(c) Let ABCD be a quadrilateral in which ∠A = ∠C and ∠B = ∠D. Prove that AB ∥ DC and that BC ∥ AD, and hence that ABCD is a parallelogram.

The next problem presents a single illustrative example of the kinds of things which we know in our bones must be true, but where the reason, or proof, may need a little thought.

Problem 160 Let ABCD be a parallelogram. Let M be the midpoint of AD and N be the midpoint of BC. Prove that MN || AB, and that MN passes through the centre of the parallelogram (where the two diagonals meet).

Problem 161 Prove that any parallelogram ABCD has the same area as the rectangle on the same base DC and “with the same height” (i.e. lying between the same two parallel lines AB and DC).

The ideas and results we have summarised up to this point provide exactly what is needed in the proof of Pythagoras’ Theorem outlined back in Chapter 1, Problem 18. They also allow us to identify two more “centres” of a given triangle ΔABC.

Problem 162 Given any triangle ΔABC, draw the line through A which is parallel to BC, the line through B which is parallel to AC, and the line through Cʹ which is parallel to AB. Let the first two constructed lines meet at C, the second and third lines meet at Aʹ, and the first and third lines meet at Bʹ.

(a) Prove that A is the midpoint of BʹCʹ, that B is the midpoint of CʹAʹ and that C is the midpoint of AʹBʹ.

(b) Conclude that the perpendicular from A to BC, the perpendicular from B to CA, and the perpendicular from C to AB all meet in a single point H. (H is called the orthocentre of ΔABC.)

Let the foot of the perpendicular from A to BC be P, the foot of the perpendicular from B to CA be Q, and the foot of the perpendicular from C to AB be R. Then ΔPQR is called the orthic triangle of ΔABC. The “circle theorems” (especially Problems 151 and 154(c)) lead us to discover that this triangle has two quite unexpected properties. As a partial preparation for one of the properties we digress slightly to introduce a classic problem.

Problem 163 My horse is tethered at H some distance away from my village V. Both H and V are on the same side of a straight river. How should I choose the shortest route to lead the horse from H to V, if I want to water the horse at the river en route?

Problem 164 Let ΔABC be an acute angled triangle.

(a) Prove that, among all possible triangles ΔPQR inscribed in ΔABC, with P on BC, Q on CA, R on AB, the orthic triangle is the one with the shortest perimeter.

(b) Suppose that the sides of ΔABC act like mirrors. A ray of light is shone along one side of the orthic triangle PQ, reflects off CA, and the reflected beam then reflects in turn off AB. Where does the ray of light next hit the side BC? (Alternatively, imagine the sides of the triangle as billiard table cushions, and explain the path followed by a ball which is projected, without spin, along PQ.)

We come next to the fourth among the standard “centres of a triangle”.

Problem 165 Given ΔABC, let L be the midpoint of the side BC. The line AL is called a median of ΔABC. (It is not at all obvious, but if we imagine the triangle as a lamina, having a uniform thickness, then ΔABC would exactly balance if placed on a knife-edge running along the line AL.) Let M be the midpoint of the side CA, so that BM is another median of ΔABC. Let G be the point where AL and BM meet.

(a) (i) Prove that ΔABL and ΔACL have equal area. Conclude that ΔABG and ΔACG have equal area.

(ii) Prove that ΔBCM and ΔBAM have equal area. Conclude that ΔBCG and ΔBAG have equal area.

(b) Let N be the midpoint of AB. Prove that CG and GN are the same straight line (i.e. that ∠CGN is a straight angle). Hence conclude that the three medians of any triangle always meet in a point G.

The point where all three medians meet is called the centroid of the triangle. For the geometry of the triangle, this is all you need to know. However, it is worth noting that the centroid is the point that would be the ‘centre of gravity’ of the triangle if the triangle is thought of as a thin lamina with a uniform distribution of mass.

Next we revisit, and reprove in the Euclidean spirit, a result that you proved in Problem 95 using coordinates - namely the Midpoint Theorem.

Problem 166 (The Midpoint Theorem) Given any triangle ΔABC, let M be the midpoint of the side AC, and let N be the midpoint of the side AB. Draw in MN and extend it beyond N to a point Mʹ such that MN = NMʹ.

(a) Prove that ΔANM = ΔBNMʹ.

(b) Conclude that BMʹ = CM and that BMʹ || CM.

(c) Conclude that MMʹBC is a parallelogram, so that CB = MMʹ. Hence MN is parallel to CB and half its length.

The Midpoint Theorem can be reworded as follows:

Given ΔAMN.

Extend AM to C such that AM = MC and extend AN to B such

that AN = NB.

Then CB || MN and CB = 2. MN.

This rewording generalizes SAS-congruence in a highly suggestive way, and points us in the direction of “SAS-similarity”.

SAS-similarity (x2): if AʹBʹ = 2 . AB, ∠BAC = ∠BʹAʹCʹ, and

AʹCʹ = 2 . AC, then

BʹCʹ = 2 . BC, ∠ABC = ∠AʹBʹCʹ, and ∠BAC = ∠BʹAʹCʹ.

Proof Extend AB to the point Bʹʹ such that ABʹʹ = AʹBʹ, and extend AC to the point Cʹʹ such that ACʹʹ = AʹCʹ. Then ΔBʹʹACʹʹ = ΔBʹAʹCʹ (by SAS-congruence), so BʹʹCʹʹ = BʹCʹ, ∠BʹʹCʹʹA = ∠BʹCʹAʹ, ∠CʹʹBʹʹA = ∠CʹBʹAʹ. By construction we have ABʹʹ = 2 . AB and ACʹʹ = 2 . AC. Hence (by the Midpoint Theorem): BʹʹCʹʹ = 2 . BC (so BʹCʹ = 2 . BC), and BC || BʹʹCʹʹ (so ∠BCA = ∠BʹʹCʹʹA and ∠CBA = ∠CʹʹBʹʹA).

∴ ∠BʹʹCʹʹA = ∠BʹCʹAʹ = ∠BCA,

and ∠CʹʹBʹʹA = ∠CʹBʹAʹ = ∠CBA. QED

The SAS-similarity ($\times 2$) interpretation of the Midpoint Theorem is like the SAS-congruence criterion in that one pair of corresponding angles in ΔBAC and ΔBʹAʹCʹ are equal, while the sides on either side of this angle in the two triangles are related; but instead of the two pairs of corresponding sides being equal, the sides of ΔBʹAʹCʹ are double the corresponding sides of ΔBAC.

In general we say that

ΔABC is similar to ΔAʹBʹCʹ (written as ΔABC ~ ΔAʹBʹCʹ) with scale-factor m if each angle of ΔAʹBʹCʹ is equal to the corresponding angle of ΔABC, and if corresponding sides are all in the same ratio:

$$\underset{\_}{A^{\prime }{B}^{\prime }}:\underset{\_}{AB}=\underset{\_}{B^{\prime }{C}^{\prime }}:\underset{\_}{BC}=\underset{\_}{C^{\prime }{A}^{\prime }}:\underset{\_}{CA}=m:1.$$

If two triangles ΔAʹBʹCʹ and ΔABC are similar, with (linear) scale factor ΔAʹBʹCʹ, then the ratio between their areas is

$$area\left(\Delta A^{\prime }{B}^{\prime }{C}^{\prime }\right):area($$ΔABC)= m 2 :1.

Two similar triangles Δ ABC and Δ A'B'C' give rise to six matching pairs:

• the three pairs of corresponding angles (which are equal in pairs), and
• the three pairs of corresponding sides (which are in the same ratio).

In the case of congruence, the congruence criteria tell us that we do not need to check all six pairs to guarantee that two triangles are congruent: these criteria guarantee that certain triples suffice. The similarity criteria guarantee much the same for similarity.

Suppose we are given triangles ΔABC, ΔAʹBʹCʹ.

AAA-similarity: If

$$\angle ABC=\angle A^{\prime }{B}^{\prime }{C}^{\prime },\angle BCA=\angle B^{\prime }{C}^{\prime }{A}^{\prime },\angle CAB=\angle C^{\prime }{A}^{\prime }{B}^{\prime },$$

then

$$\underset{\_}{A^{\prime }{B}^{\prime }}:\underset{\_}{AB}=\underset{\_}{B^{\prime }{C}^{\prime }}:\underset{\_}{BC}=\underset{\_}{C^{\prime }{A}^{\prime }}:\underset{\_}{CA},$$

so the two triangles are similar.

SSS-similarity: If

$$\underset{\_}{A^{\prime }{B}^{\prime }}:\underset{\_}{AB}=\underset{\_}{B^{\prime }{C}^{\prime }}:\underset{\_}{BC}=\underset{\_}{C^{\prime }{A}^{\prime }}:\underset{\_}{CA},$$

then

$$\angle ABC=\angle A^{\prime }{B}^{\prime }{C}^{\prime },\angle BCA=\angle B^{\prime }{C}^{\prime }{A}^{\prime },\angle CAB=\angle C^{\prime }{A}^{\prime }{B}^{\prime },$$

so the two triangles are similar.

SAS-similarity: If

$$\underset{\_}{A^{\prime }{B}^{\prime }}:\underset{\_}{AB}=\underset{\_}{A^{\prime }{C}^{\prime }}:\underset{\_}{AC}=m:1$$

and

$$\angle B^{\prime }{A}^{\prime }{C}^{\prime }=\angle BAC,$$

then

$$\underset{\_}{B^{\prime }{C}^{\prime }}:\underset{\_}{BC}=\underset{\_}{A^{\prime }{B}^{\prime }}:\underset{\_}{AB}={\underset{\_}{A}}^{\prime }{C}^{\prime }:\underset{\_}{AC}$$

and

$$\angle A^{\prime }{B}^{\prime }{C}^{\prime }=\angle ABC,\angle B^{\prime }{C}^{\prime }{A}^{\prime }=\angle BCA,$$

so the two triangles are similar.

Our rewording of the Midpoint Theorem gave rise to a version of the third of these criteria, with m = 2.

AAA-similarity in right angled triangles is what makes trigonometry possible. Suppose that two triangles ΔABC,ΔAʹBʹCʹ have right angles at A and at Aʹ. If ∠ABC,∠AʹBʹCʹ, then (since the angles in each triangle add to two right angles) we also have ∠BCA,∠BʹCʹAʹ. It then follows (from AAA-similarity) that

$$\underset{\_}{A^{\prime }{B}^{\prime }}:\underset{\_}{AB}=\underset{\_}{B^{\prime }{C}^{\prime }}:\underset{\_}{BC}=\underset{\_}{C^{\prime }{A}^{\prime }}:\underset{\_}{CA},$$

so the trig ratio in ΔABC

$$\sin B=\frac{\underset{\_}{AC}}{\underset{\_}{BC}}$$

has the same value as the corresponding ratio in ΔAʹBʹCʹ

$$\sin B^{\prime }=\frac{\underset{\_}{A^{\prime }{C}^{\prime }}}{\underset{\_}{B^{\prime }{C}^{\prime }}}.$$

Hence this ratio depends only on the angle B, and not on the triangle in which it occurs. The same holds for cos ∠B and for tan ∠B.

The art of solving geometry problems often depends on looking for, and identifying, similar triangles hidden in a complicated configuration. As an introduction to this, we focus on three classic properties involving circles, where the figures are sufficiently simple that similar triangles should be fairly easy to find.

Problem 167 The point P lies outside a circle. The tangent from P touches the circle at T, and a secant from P cuts the circle at A and at B. Prove that $\underset{\_}{PA}\times \underset{\_}{PB}={\underset{\_}{PT}}^2$ .

Problem 168 The point P lies outside a circle. Two secants from P meet the circle at A, B and at C, D respectively. Prove in two different ways that

$$\underset{\_}{PA}\times \underset{\_}{PB}=\underset{\_}{PC}\times \underset{\_}{PD}.$$

Problem 169 The point P lies inside a circle. Two secants from P meet the circle at A, B and at C, D respectively. Prove in two different ways that

$$\underset{\_}{PA}\times \underset{\_}{PB}=\underset{\_}{PC}\times \underset{\_}{PD}.$$

We end our summary of the foundations of Euclidean geometry by deriving the familiar formula for the area of a trapezium and its 3-dimensional analogue, and a formulation of the similarity criteria which is often attributed to Thales (Greek 6^th century BC).

Problem 170 Let ABCD be a trapezium with AB || DC, in which AB has length a and DC has length b.

(a) Let M be the midpoint of AD and let N be the midpoint of BC. Prove that MN || AB and find the length of MN.

(b) If the perpendicular distance between AB and DC is d, find the area of the trapezium ABCD.

Problem 171 A pyramid ABCDE, with apex A and square base BCDE of side length b, is cut parallel to the base at height d above the base, leaving a frustum of a pyramid, with square upper face of side length a. Find a formula for the volume of the resulting solid (in terms of a, b, and d).

The following general result allows us to use “equality of ratios of line segments” whenever we have three parallel lines (without first having to conjure up similar triangles).

Problem 172 (Thales’ Theorem) The lines AA' and BB' are parallel. The point C lies on the line AB, and C' lies on the line A'B' such that CCʹ || BBʹ. Prove that AB : BC = AʹBʹ : BʹCʹ.

Under certain conditions, the similarity criteria guarantee the equality of ratios of sides of two triangles. Thales’ Theorem extends this “equality of ratios” to line segments which arise whenever two lines cross three parallel lines. One of the simplest, but most far-reaching, applications of this result is the tie-up between geometry and algebra which lies behind ruler and compasses constructions, and which underpins Descartes’ (1596-1650) re-formulation of geometry in terms of coordinates (see Problem 173).

Thales (c. 620-c. 546 BC) was part of the flowering of Greek thought having its roots in Milesia (in the south west of Asia Minor, or modern Turkey). Thales seems to have been interested in almost everything - philosophy, astronomy, politics, and also geometry. In Britain his name is usually attached to the fact that the angle subtended by a diameter is a right angle. On the continent, his name is more strongly attached to the result in Problem 172. His precise contribution to geometry is unclear - but he seems to have played a significant role in kick-starting what became (300 years later) the polished version of Greek mathematics that we know today.

Thales’ contributions in other spheres were perhaps even more significant than in geometry. He seems to have been among the first to try to “explain” phenomena in reductionist terms - identifying “water” as the single “element”, or first principle, from which all substances are derived. Anaximenes (c. 586-c. 526 BC) later argued in favour of “air” as the first principle. These two elements, together with “fire” and “earth”, were generally accepted as the four Greek “elements” - each of which was supposed to contribute to the construction of observed matter and change in different ways.

Problem 173 To define “length”, we must first decide which line segment is deemed to have unit length. So suppose we are given line segments XY of length 1, AB of length a, (i.e. AB : XY = a : 1), and CD of length b.

(a) Use Problem 137 to construct a segment of length a + b, and if $a⩾b$, a segment of length a — b.

(b) Show how to construct a line segment of length ab and a segment of length $\frac{a}{b}$ .

(c) Show how to construct a line segment of length $\sqrt{a}$ .