In his thirteen books of Elements, Euclid developed long sequences of propositions, each relying on the previous ones. According to Morrow, p. xxii, we have very little exact information about the author of this remarkable work, which lays the foundations for various fields of mathematics, including plane geometry, the theory of proportion, number theory, incommensurable magnitudes, and solid geometry. The success of Euclid's book came from his skillful selection of the propositions and in their arrangement into a logical sequence, following deductively from a small handful of initial assumptions (Eves, p. 144). Euclid provides us with some definitions, axioms, and postulates before he attempts to prove his propositions. Euclid's propositions are ordered in such a way that each proposition is only used by future propositions and never by any previous ones. In Appendix A, there is a chart of all the propositions from Book I that illustrates this. Proposition 47 in Book I is probably Euclid's most famous proposition: the "Pythagorean Theorem". I will illustrate in detail how Euclid proves Proposition 47 by showing how he uses his previous propositions in proving it. I will discuss all the propositions (and their proofs) that are used directly in the proof of Proposition 47, but I will not discuss any other propositions nor proofs that are not used directly.
Before we can begin, we need to state Proposition 47: In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle. For example if a right-angled triangle ABC has angle BAC as the right angle, then the square on side BC equals the sum of the squares of sides BA and AC.
According to Euclid, the first step of the proof requires us to construct (or "describe") squares on all three sides of the triangle, and in order to do this, we need to use his Proposition 46, which says: On a given straight line to describe a square. But notice how Proposition 46 says we describe a square, instead of constructing one. In his previous propositions, he constructs a triangle, instead of describing it (e.g. Proposition 1: On a given finite straight line to construct an equilateral triangle). Proclus and Heath both take notice of this, and they both give a brief comment on the matter. Proclus says: Since the triangle is put together by many parts, it must be constructed, while the square is generated from one of its sides, and must be described (Proclus, p. 423). Heath on the other hand says: The triangle (or angle) is, so to say, pieced together, while the describing of a square on a given straight line is the making of a figure "from" one side, and corresponds to the multiplication of the number representing the side by itself (Heath, p. 428). Even after reading these two authors' comments, I am still not sure why Euclid chooses these two separate terms. Perhaps Euclid views the triangle and the square as two uniquely unrelated shapes and so he decides to give them their own individual terms "construct" and "describe" for creating them. Since the reason for the two terms are irrelevant to my main topic, I shall not discuss this any further, and will instead continue on with the proof of Proposition 47.
Let us consider the structure of the construction and proof carried out in Proposition 46, taking note of the earlier propositions used. In order to describe a square, we would first need to let AB be a given straight line and then to describe a square on AB. First we would need to draw a line AC at right angles to the straight line AB from the point A on it. This first step comes from Euclid's proof of Proposition 11: To draw a straight line at right angles to a given straight line from a given point on it. Next make AD equal to AB, which comes from Proposition 3: Given two unequal straight lines, to cut off from the greater a straight line equal to the less. Now draw DE through the point D parallel to AB, and draw BE through the point B parallel to AD. This step comes from Proposition 31: Through a given point to draw a straight line parallel to a given straight line, which proved later on when it is needed again in Proposition 47.
Now the figure ADEB is a parallelogram. By using Proposition 34: In parallelogrammic areas the opposite sides and angles equal one another, and the diameter bisects the areas, as well as some intuition by inspection of the figure, it can be proved that parallelogram ADEB has sides that are all equilateral, and that all the angles are all right-angled. And since this is Euclid's definition of a square, this parallelogram is a square described on the straight line AB. And this concludes the proof of Proposition 46.
Let us continue with the proof of Proposition 47 by describing a square on all three sides of the triangle, as shown in the figure above. We can now describe the square BDEC onto BC, the squares GB and HC onto BA and AC. Now by using Proposition 31 and Postulate 1 we can draw a line AL through A parallel to either side BD or CE, and join AD and FC. This is illustrated in the next figure.
Since Proposition 31 is now being invoked, I will now suspend the proof of Proposition 47 and give the proof for Proposition 31: Through a given point to draw a straight line parallel to a given straight line. Let us start by using the figure above and letting A be a given point, and BC a given straight line. It is required to draw a straight line through the point A parallel to the straight line BC. This is from Postulate 1: To draw a straight line from any point to any point. Join AD. Now from Proposition 23: To construct a rectilinear angle equal to a given rectilinear angle on a given straight line and at a point on it, we can construct the angle DAE equal to the angle ADC on the straight line DA and at the point A on it. Next we can produce the straight line AF in a straight line with EA, which comes from Postulate 2: To produce a finite straight line continuously in a straight line.
Since the straight line AD falling on the two straight lines BC and EF makes the alternate angles EAD and ADC equal to one another, therefore EAF is parallel to BC. This comes from Proposition 27: If a straight line falling on two straight lines makes the alternate angles equal to one another, then the straight lines are parallel to one another. Therefore the straight line EAF has been drawn through the given point A parallel to the given straight line BC. And this concludes the proof of Proposition 31.
Resuming the proof of Proposition 47, since each of the angles BAC and BAG are right, it follows that with a straight line BA, and at the point A on it, the two straight lines AC and AG not lying on the same side make the adjacent angles equal to two right angles, therefore CA is in a straight line with AG. And for the same reason BA is also in a straight line with AH. This is proved by Proposition 14: If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines are in a straight line with one another. We must pause here in the proof of Proposition 47 in order to examine the proof of Proposition 14.
The proof of Proposition 14 starts by drawing any straight line AB, and at the point B on it, let the two straight lines BC and BD not lying on the same side make the sum of the adjacent angles ABC and ABD equal to two right angles. Now we must prove that BD is in a straight line with CB.
Let us first assume that if BD is not in a straight line with BC, we can then produce BE in a straight line with CB, as stated in Postulate 2. Since the straight line AB stands on the straight line CBE, therefore the sum of the angles ABC and ABE equals two right angles; from Proposition 13: If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles. But the sum of the angles ABC and ABD also equals two right angles, which comes from Postulate 4: That all right angles equal one another. And from Axiom 1: Things that equal the same thing also equal one another; therefore the sum of the angles CBA and ABE equals the sum of the angles CBA and ABD.
Now using Axiom 3: If equals are subtracted from equals, then the remainders are equal, we can subtract the angle CBA from each. Then the remaining angle ABE equals the remaining angle ABD, the less equals the greater, which is impossible. Therefore BE is not in a straight line with CB, which means that CB is in a straight line with BD. Thus Proposition 14 is proved.
Continuing on with Proposition 47, from the definition of a square, since the angle DBC equals the angle FBA, and each is right. We can add the angle ABC to each, therefore the whole angle DBA equals the whole angle FBC by using Postulate 4 and Axiom 2. Postulate 4 says: If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides and Axiom 2 says: If equals are added to equals, then the wholes are equal.
Since DB equals BC, and FB equals BA, the two sides AB and BD equal the two sides FB and BC respectively, and the angle ABD equals the angle FBC, therefore the base equals the base FC, and the triangle ABD equals the triangle FBC. This comes from the definition of a square and Proposition 4: If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, then they also have the base equal to the base, the triangle equals the triangle, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides.
To prove Proposition 4, let ABC and DEF be two triangles having the two sides AB and AC equal to the two sides DE and DF respectively, namely AB equal to DE and AC equal to DF, and the angle BAC equal to the angle EDF. Now we need to prove that the base BC also equals the base EF, the triangle ABC equals the triangle DEF, and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides, that is, the angle ABC equals the angle DEF, and the angle ACB equals the angle DFE.
If the triangle ABC is superposed on the triangle DEF and if the point A is placed on the point D and the straight line AB on DE, then the point B also coincides with E, because AB equals DE. Again, AB coinciding with DE, the straight line AC also coincides with DF, because the angle BAC equals the angle EDF. Hence the point C also coincides with the point F, because AC again equals DF.
Using Axiom 4: Things which coincide with one another equal one another, we can say that B also coincides with E, hence the base BC coincides with the base EF and equals it. Thus the whole triangle ABC coincides with the whole triangle DEF and equals it. And the remaining angles also coincide with the remaining angles and equal them, the angle ABC equals the angle DEF, and the angle ACB equals the angle DFE. This concludes the proof of Proposition 4.
Continuing on with Proposition 47, the parallelogram BL is double the triangle ABD, since they have the same base BD and are in the same parallels BD and AL. And the square GB is double the triangle FBC, for they again have the same base FB and are in the same parallels FB and GC. This is proved in Proposition 41: If a parallelogram has the same base with a triangle and is in the same parallels, then the parallelogram is double the triangle. Therefore the parallelogram BL also equals the square GB.
Now let us prove the final proposition that is needed to finish proving Proposition 47. The proof of Proposition 41 goes as follows: Let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC and AE. Now lets prove that the parallelogram ABCD is double the triangle BEC. From Postulate 1, join AC. Then from Proposition 37: Triangles which are on the same base and in the same parallels equal one another; the triangle ABC equals the triangle EBC, since it is on the same base BC with it and in the same parallels BC and AE. But the parallelogram ABCD is double the triangle ABC, since the diameter AC bisects it, so that from Proposition 34; the parallelogram ABCD is also double the triangle EBC. And this concludes the proof of Proposition 41.
Now we can finally complete the proof of Proposition 47 and verify the Pythagorean Theorem. All that is needed to be done is to carry out similar steps if lines AE and BK are joined. The parallelogram CL can also be proved equal to square HC, similarly. Therefore from Axiom 2: If equals are added to equals, then the wholes are equal, the whole square BDEC equals the sum of the two squares GB and HC. Since the square BDEC is described on BC, and the squares GB and HC are described on BA and AC, the square on the side BC equals the sum of the squares on BA and AC.
Therefore in right-angled triangles the square on the side subtending the right angle equals the sum of the squares on the sides containing the right angle. In Euclid's Book I, Proposition 48, he proves its converse also holds true. Proposition 48 says that if in a triangle the square on one of the sides is equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.
This concludes our analysis of the proof of Proposition 47. This analysis demonstrates how Euclid uses previous propositions that he has proven before constructing this proof. These propositions were placed in such a logical order such that no proposition uses a proposition that was not already proven before it. Although this is the first solid proof of the Pythagorean Theorem that has come down to us, Euclid laid hold of a theorem even more general than this and secured it by irrefutable scientific arguments. In his sixth book, Euclid proves in Proposition 31: In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly situated figures described on the sides about the right angle. This requires Eudoxos' theory of proportions, developed in Book V.
Here is a table that lists all the sequences of propositions (in bold), each relying on the previous (in italics). Note that Propositions 1 & 4 are the only two propositions that rely only on the postulates and axioms. 1
|1|| ||13||11||25||4, 24||37||31, 34, 35|
|2||1||14||11||26||3, 4, 16||38||31, 34, 36|
|3||1||15||11||27||3, 4, 10, 15||39||31, 34, 36|
|4|| ||16||3, 4, 10, 15||28||13, 15, 27||40||31, 38|
|5||3, 4||17||13, 16||29||13, 15, 27||41||34, 37|
|6||3, 4||18||3, 5, 16||30||23, 27||42||10, 23, 31, 38, 41|
|7||5||19||5, 18||31||13, 29, 31||43||4, 26, 29|
|8||5||20||3, 5, 19||32||4, 27, 29||44||15, 29, 31, 42, 43|
|9||1, 3, 8||21||16, 20||33||4, 26, 29||45||14, 29, 39, 33, 34, 42, 44|
|10||1, 4, 9||22||3, 20||34||4, 29, 34||46||3, 11, 29, 31, 34|
|11||1, 3, 8||23||8, 22||35||33, 34, 35||47||4, 14, 31, 41, 46|
|12||8, 10||24||3, 4, 5, 19, 23||36||31, 34, 35||48||3, 8, 11, 47|