Euclid’s Elements

Geometry begins by agreeing on what its words mean. Think of a point as a location without size, a segment as the straight connection between two endpoints, and a circle as the points in a plane at one fixed distance from a center. The ink on a page represents these ideal objects; it is not the object itself.

A definition fixes a meaning. A postulate supplies a starting permission or assumption. A proposition is a claim you must establish from those starting points and earlier results. Euclid allows us to join points, extend a segment, and draw a circle with a given center and radius. He also assumes equality of right angles and a condition about when lines meet.

A proof explains why a conclusion follows. A drawing helps you discover a route, but measuring the drawing cannot establish a claim about every possible triangle. Throughout this course, AB means the length of segment AB when it appears in an equation; ∠ABC means the angle with vertex B.

Euclid’s common notions let us carry information from one step to the next: quantities equal to one quantity are equal to each other; adding equal amounts preserves equality; subtracting equal amounts preserves equality. Coinciding things are equal, and a whole exceeds a proper part in the finite magnitudes under discussion.

Always ask: equal in what respect? Two rectangles can have equal areas but different side lengths. Compare like quantities: a length with a length, an area with an area. An equation is only meaningful when its terms refer to compatible quantities.

Given distinct points A and B, draw the circle centered at A through B. Draw the circle centered at B through A. Choose an intersection C and join AC and BC. The circles are working tools: each carries an equality into the construction.

AC = AB because both are radii of the first circle. BC = BA because both are radii of the second. AB and BA are the same segment. Therefore AC = BC by Common Notion 1. All three sides are equal, so ABC is equilateral.

One subtlety matters: Euclid treats the circles’ intersection as available without an explicit continuity postulate. We assume the usual continuous Euclidean plane here. Good proof reading includes noticing assumptions, even in a famous text.

Congruent triangles have the same shape and size, even if one is turned over or rotated. Match vertices before using the result: if ABC corresponds to DEF, then A ↔ D, B ↔ E, and C ↔ F.

Side–side–side (SSS) uses three pairs of equal sides. Side–angle–side (SAS, I.4) uses two pairs of equal sides and the equal angle between them. Once congruence is established, corresponding angles and sides agree.

When two triangles share a side, that side is equal to itself. Do not overlook this free piece of information. Equal angles alone give similarity, not equal size. Two sides and an angle that is not between them do not in general determine a unique triangle.

Let A be the vertex of the angle. Take D on one ray, and E on the other so AD = AE. Join DE. Construct an equilateral triangle DEF on the side of DE away from A, and join AF. For an ordinary angle between 0° and 180°, this places AF inside it.

Now compare triangles ADF and AEF. AD = AE by construction; DF = EF because DEF is equilateral; AF = AF because it is shared. By SSS the angles DAF and EAF are equal. Therefore AF bisects the original angle.

Notice the pattern: construct something whose equalities you control, find two triangles, and use those equalities to get the angle you need.

Use five headings: Given; To prove; Construction, if needed; Reasons; Conclusion. Work backward to discover a route: what would be enough to establish the goal? Then write forward, so every step uses something already available.

Here is a modern short proof of the base-angle result, using SSS rather than Euclid’s original I.5 argument. Given AB = AC, compare ABC with ACB. AB = AC is given, AC = AB is the same equality reversed, and BC = CB is the same segment. SSS matches B with C, so ∠ABC = ∠ACB.

This proof is valid after SSS (I.8) is available. It would be circular to insert it earlier into a development that needs I.5 to prove I.8. Validity depends not only on true statements, but on which results are already established.

Parallel lines lie in the same plane and do not meet however far extended. A transversal is a line crossing two other lines. In Euclidean geometry, parallel lines give equal alternate interior angles and equal corresponding angles.

Euclid’s fifth postulate says, in effect: when a transversal makes the interior angles on one side total less than two right angles, the two lines meet on that side if extended. A familiar equivalent formulation, with the usual other assumptions, is that through a point outside a line exactly one parallel can be drawn.

I.29 uses that postulate. Do not silently replace “these angles look alike” with “these lines are parallel.” A justified parallel construction or an earlier result is needed. Changing the parallel assumption leads to different geometries; the conclusions depend on the system.

Take triangle ABC. Through C construct a line parallel to AB, and extend BC past C to D. Let E lie on the new parallel on the side used to split exterior angle ACD.

By the parallel-angle result, ∠ACE = ∠CAB and ∠ECD = ∠ABC. The two angles at C together make the exterior angle ACD. Add ∠BCA. The angles ACD and BCA form a straight angle, so the three interior angles of ABC total two right angles, or 180°.

The useful new line was not in the givens. Adding a justified auxiliary construction is one of the main skills of proof. The result applies to triangles in the Euclidean plane, not automatically to triangles drawn on a sphere.

For a right triangle with legs a and b and hypotenuse c, the area of the square on c equals the combined areas of the squares on a and b. The squares in a² + b² = c² are areas, not a decorative way of writing lengths.

A modern rearrangement proof: arrange four copies of the triangle inside a square of side a + b, leaving a central quadrilateral whose sides are all c. The acute angles of a right triangle sum to 90°, so every corner of the central quadrilateral is a right angle. It is a square. Its area is (a + b)² − 4(ab/2) = a² + b², and also c².

This is a teaching proof of the same theorem, not Euclid’s I.47 proof. His proof compares triangles and rectangles in squares erected on the sides. Read it after you are comfortable with equality of areas.

Split a segment into positive lengths a and b. Make a square of side a + b, then draw one horizontal and one vertical cut at the split. The four regions are a square of area a², a square of area b², and two rectangles each of area ab.

The pieces cover the original square without overlap or gaps. Therefore (a + b)² = a² + 2ab + b². This notation is a modern translation of a geometric relationship in Book II, not Euclid writing symbolic algebra.

An area proof works by accounting for every piece. You may rearrange equal pieces, but you must establish that the original and new arrangements cover the claimed regions exactly.

Let AB be a diameter of a circle with center O, and let C be any other point on the circle. Join OC, AC, and BC. Since OA = OC and OB = OC, triangles AOC and BOC are isosceles.

Write α for ∠OAC = ∠ACO and β for ∠OBC = ∠BCO. Because O lies on AB, the large triangle ABC has angles α, β, and α + β. Their sum is 180°, so 2(α + β) = 180°. Therefore ∠ACB = 90°.

This is the semicircle case of III.31. The diameter hypothesis does real work: an arbitrary chord does not give this conclusion. A boundary point C cannot be one of the endpoints A or B.

Book IV asks us to place regular figures inside or around circles. “Inscribed” means the vertices lie on the circle; “circumscribed” means the sides touch the circle. A regular polygon has equal sides and equal interior angles.

For a regular hexagon inscribed in a circle, each side has the same length as the radius. Think of six triangles meeting at the center: each has three radius-length sides, so it is equilateral. Each central angle is 60°, and six such angles make a full turn.

With a compass set to the radius, step equal chords around the circle and join consecutive points. The angle argument explains why six steps close the figure. Construction requires an exact reason for closure, not merely a convincing final dot.

A ratio compares two positive magnitudes of the same kind. Equal ratios mean the comparisons scale in the same way, not that the magnitudes themselves are equal. Thus 2:3 and 4:6 agree, although 2 does not equal 4.

Book V does not require every pair of lengths to be whole-number multiples of a shared unit. Its test for a:b = c:d compares m·a with n·b and m·c with n·d for every pair of positive integers m and n. In both comparisons, less than, equal to, or greater than must agree.

“Every” is crucial. Passing a few comparisons does not establish the definition. Modern fraction arithmetic handles many examples quickly, but the general theory also accommodates incommensurable lengths.

Similar triangles have equal corresponding angles and proportional corresponding sides. Unlike congruence, similarity permits a change of scale. Write the correspondence first, then choose ratios in the same order.

If all side lengths grow by a factor k, areas grow by k². Doubling a square’s side produces four copies of its original area, not two. Scale has different effects on length and area.

A shadow measurement translates a physical situation into a geometry problem. It works when the objects are upright, the ground is level, and sunlight gives effectively parallel rays at the same time. The mathematical conclusion is conditional on that model.

Books VII–IX develop arithmetic. A number measures another when it divides it exactly. Euclid finds a greatest common measure by repeatedly removing the smaller magnitude from the larger. The familiar division-with-remainder algorithm compresses those subtractions.

For 48 and 18: 48 = 2·18 + 12; 18 = 1·12 + 6; 12 = 2·6 + 0. The last nonzero remainder is 6. Why does this work? If a = qb + r, any common divisor of a and b divides r = a − qb. Conversely, any common divisor of b and r divides a = qb + r. Each step preserves exactly the common divisors.

The positive remainders strictly decrease, so this process terminates. Book VIII goes on to continued proportions, such as 2:4 = 4:8. “What stays unchanged?” and “why must it stop?” are powerful questions about any algorithm.

Suppose someone offers a finite list of primes p₁, …, pₙ. Form N = p₁·…·pₙ + 1. Dividing N by any listed prime leaves remainder 1, so none of them divides N.

Because N > 1, it has a prime divisor. That divisor is outside the list. Therefore every finite list misses a prime. This modern product-plus-one presentation expresses the argument behind IX.20; Euclid states the result as finding more primes than any assigned collection.

Crucially, N need not itself be prime. The argument only needs a prime divisor of N. Euclid’s earlier arithmetic provides the fact that a composite number has a prime divisor (VII.31). This is a dependency, not a fact to leave unexplained.

Two lengths are commensurable if both are whole-number multiples of one common length. A unit square’s side and diagonal are not: Pythagoras makes their ratio √2. Book X studies incommensurable magnitudes in much greater detail than this introductory example.

Here is a modern parity proof. Suppose √2 = m/n for positive integers m and n with no common divisor greater than 1. Squaring gives m² = 2n². Thus m² is even, so m is even: the square of an odd integer is odd. Write m = 2k. Substitution gives n² = 2k², so n is even too.

But then m and n share 2, contradicting the reduced-fraction assumption. Therefore no such fraction exists. The proof does not say the diagonal is unknowable or immeasurable in practice; it says no rational number is its exact ratio to the side.

Book XI establishes relationships among lines and planes in space. A line perpendicular to one line in a plane is not necessarily perpendicular to the whole plane. Moving from a drawing to a spatial claim requires extra conditions.

Book XII compares areas and volumes using increasingly fine inscribed figures and arguments that rule out an unequal remainder. This method of exhaustion is an ancestor of limit reasoning, not modern integral notation. For example, XII.7 yields that a triangular-based pyramid has one third the volume of a prism with the same base and height.

Book XIII constructs the five convex regular solids: tetrahedron, cube, octahedron, dodecahedron, and icosahedron. At a vertex, the face angles must total less than 360°. This angle budget narrows the possibilities; Euclid’s constructions supply the existence arguments. A list of plausible candidates alone is not a complete classification proof.

The Ethics presents definitions and axioms, then propositions with demonstrations, corollaries, and scholia. A corollary draws a further consequence; a scholium offers discussion or explanation. The arrangement lets a reader inspect how later claims depend on earlier ones.

For example, Ethics I, proposition 3 argues that things with nothing in common cannot cause one another. Its demonstration appeals to axiom 5, about understanding things through one another, and axiom 4, about understanding an effect through its cause. Trace those links before deciding whether you accept the conclusion.

Euclid practice teaches you to ask: what is given, which definition is active, and why does this step follow? It does not make Spinoza’s metaphysics an uncontested mathematical result. You can understand the logical structure while questioning a definition, axiom, or inference.

In the preface to Part III, Spinoza proposes to examine emotions within nature’s order rather than treating people as exempt from it. A useful contemporary reading practice is to look for causes and conditions instead of stopping at praise or blame. The exercises here are our applications, not statements that Spinoza wrote about modern schedules or workplaces.

Try three labels. DEDUCTION: if the accepted premises hold, the conclusion follows. OBSERVATION: a claim about the world that needs evidence and may be revised. VALUE: something you choose to prioritize. Mixing them makes a preference look like a fact, or a guess look like a necessity.

Example: “If I reserve 45 of my 60 free minutes, 15 remain” is arithmetic. “Turning off notifications will help me finish” is an empirical hypothesis. “Finishing this matters more than checking messages” expresses a priority. A sound plan names all three and tests the uncertain part.

Final construction: given distinct A and B, construct equilateral ABC. Bisect ∠ACB with a ray meeting AB at D. Prove AD = DB. You already have the tools: I.1, I.9, and SAS (I.4).

Plan backward: equal corresponding sides would follow from congruent triangles ACD and BCD. Plan forward: AC = BC because ABC is equilateral; CD is shared; ∠ACD = ∠BCD by angle bisection. These are the included angles, so SAS gives the congruence and AD = DB.

Write the proof in the notebook below before looking at its model answer. Name D’s position on AB as part of the construction in the usual Euclidean plane. Check each hypothesis of SAS and finish with the original goal. Then try a second proof of your own choosing: a triangle angle sum, the semicircle theorem, or the irrationality of √2.