Clavius' Edition

Unlike medieval Latin script, 17th-century Latin is much easier to read. I used a combination of AI tools to translate Clavius'  Book I, Definition 1, and Book II, Proposition 4  from Latin to English.  What is striking about this text is that it offers more explanation than most other editions and probably explains the popularity of Clavius' edition.  This edition is over 750 pages long.

Book 1, Definition 1: 

In Book 1 of Euclid, the single phrase is: 

Punctum est cuius pars nulla est. 

In English: A point is that which has no part. 

Most editions leave it at that, but not Clavius; he writes the following extended commentary, first introducing the Book, then on the nature of space.

This whole first book is laid out so as to give us the origins and properties of triangles — both with respect to their angles and with respect to their sides — which it sometimes compares with one another, and sometimes examines and considers each one on its own. For at times it studies the angles of a triangle from its sides, and at other times it investigates the sides from the angles with respect to equality and inequality. And it makes the same inquiry by various methods when two triangles are compared with each other. Next it opens up for us the properties of parallels, and undertakes the study of parallelograms — both among themselves and also as they are compared with triangles set up between the same parallels.

But so that Euclid may carry all this out more correctly and conveniently, he teaches the division of a rectilinear angle and of a straight line into equal parts, the construction of a perpendicular line, how one angle may be made equal to another, and other things of this kind. And so, to sum up the whole matter in a word: in the first book there are set forth — following Proclus's view — the foremost and principal of rectilinear figures, namely triangles and parallelograms.

Before everything else, however, Euclid, in the manner of mathematicians, begins the proposed subject from first principles, making his start from definitions, the first of which explains the point, teaching that that is called a point, in continuous quantity, which has no parts. And this definition will be perceived more clearly and easily if we first understand that continuous quantity has three kinds of parts: some according to length, others according to breadth, and others according to depth or height. Not every quantity, however, has all these parts: some have only the single kind, according to length; some have two kinds, so that to the former it adds the parts of breadth as well; and some, finally, contain besides these two kinds a third as well, of height or depth. For every continuous quantity is either long only, or at once long and broad, or long, broad, and deep. Nor can any quantified thing have another dimension, as Ptolemy rightly demonstrated in his little book On the Analemma — recently restored to its former dignity by the labor of Federico Commandino of Urbino — and also, as Simplicius says, in his little book On Dimension, which indeed, so far as I know, has not yet been printed.

And so, whatever exists in continuous quantity, that is, in magnitude, and is understood to be without any part, such that it is conceived as neither long nor broad nor deep — that is called by Euclid, and by geometers, a point. No example of this can be found among material things, unless you should wish to take the tip of some very sharp needle as expressing the likeness of a point; which, however, is not true in every respect, since that tip can be divided and cut infinitely, whereas a point must be reckoned altogether indivisible. Finally, that must be conceived to be a point in magnitude which unity is in number, and which the instant is in time. For these too are to be conceived as indivisible.

One thing worth noting of historical interest: Clavius mentions a quote from Simplicius (480-560 AD) who cites in his commentary on Aristotle's De caelo, from Ptolemy's lost On Dimension  where he argued there can be no fourth dimension. Clavius remarks that the book On Dimension "so far as I know, has not yet been printed." In fact, it was never printed because the book On Dimension was never found and appears to be lost to us.

Proposition 4

Proposition 4 proves the relation AB^2 = AC^2 + CB^2 + 2(AC ·. CB). The proof is long-winded, but the approach is quite straightforward. 

The basis of the proof is to break up a square into four rectangles; this is shown in the figure below. The key to understanding the approach is to realize that, by construction, the two yellow rectangles are the same size.  The proposition simply says that the area of the big square (AB) is the sum of the four smaller rectangles. 

The two identical yellow rectangles is where the 2 (AC . CB) comes from, noting that CB is the width of the second yellow rectangle.  Note that Clavius' figure is inverted compared to what I use. The 888 AD d'Orville manuscript uses the orientation I use here, as do most modern editions. The 12th Greek to Latin edition has it inverted like Clavius, while Adelard's edition has it as we show below.  

Book II, Proposition 4.

If a straight line is divided in any manner, the square on the whole line is equal to the squares on the two segments together with twice the rectangle contained by the segments. 

The straight line AB, divided at the point C.

I say that the square on the whole straight line AB is equal to the squares on the segments AC and CB, together with twice the rectangle contained by the segments AC and CB.

Let the square AD be described upon AB, and let the diagonal BE be drawn. Next from C let CF be drawn parallel to the straight line BD, cutting the diagonal at G. Through this point let HI again be drawn parallel to the straight line AB. Thus the square AD is divided into four parallelograms.

Since in triangle ABE the two sides AB and AE are equal, the two angles ABE and AEB are also equal. But the angles ABE, AEB, and BAE of triangle **ABE** are together equal to two right angles; and BAE is a right angle. Therefore the remaining angles ABE and AEB are each half a right angle.

By the same reasoning the angles DBE and DEB are also half-right angles. This also follows from what we proved in Proposition 4 of Book I; for, since BE is the diagonal of the square, it bisects the right angles ABD and AED.

Therefore the three angles of triangle EFG are together equal to two right angles. Since angle EFG is a right angle, being equal to the right angle at D by reason of parallel lines, and angle FEG is half a right angle, the remaining angle EGF is likewise half a right angle. Hence angle FEG equals angle EGF.

Therefore the sides EF and FG are equal; and since these are equal respectively to the opposite sides GH and HE, the figure FH is a parallelogram, indeed a square, because all its sides are equal and all its angles are right angles.

By the same reasoning CI is also a square.

Since, therefore, CI and FH are the squares on the segments AC and CB, while HG is equal to AC, and the rectangles AG and DG are contained by the segments AC and CB (because CG and GI are each equal to CB, and FG, equal to GH, is equal to AC), it follows that, since the square AD is equal to the squares CI and FH together with the rectangles AG and DG, the square AD, that is, the square on the whole line AB, is equal to the squares on the segments AC and CB, together with twice the rectangle contained by those same segments AC and CB.

Therefore, if a straight line is divided into any two parts, the square described upon the whole line is equal to the squares on the two parts together with twice the rectangle contained by those parts.

Which was to be demonstrated.

Alternatively

Since the straight line AB is divided at C, the square of the whole AB will be equal to the rectangles contained under the whole AB and the segments AC, CB.

But the rectangle contained under AB and AC is equal to the rectangle contained under AC, CB, and the square of the segment AC.

Likewise, the rectangle contained under AB, CB is equal to the rectangle contained under CB, AC, and the square of the segment CB.

Therefore, the square of the straight line AB is equal to the squares of the segments AC, CB, and to the rectangles under AC, CB and under CB, AC.

Which is what was proposed.

Modern Mathematical Meaning

This passage provides a purely algebraic-style alternative proof for the identity (a+b)^2 = a^2 + b^2 + 2ab, using earlier propositions from Euclid Book II:

   1. AB^2 = (AB . AC) + (AB . CB) (The square of a whole line equals the rectangles formed by the whole line and its parts).

   2. AB . AC = AC^2 + (AC . CB) (A rectangle formed by a whole line and one part equals the square of that part plus the rectangle of the two parts).

   3. AB . CB = CB^2 + (CB . AC) (The same rule applied to the other part).

   4. Substituting steps 2 and 3 into step 1 yields the final proof: AB^2 = AC^2 + CB^2 + 2(AC . CB).

Corollary

From this it is manifest that parallelograms about the diameter of a square are squares.

This is evident from the previous demonstration of this theorem, in which it was shown that the rectangles CI and FH, which are about the diameter BE, are squares. For in all other squares, the same demonstration will apply.

However, this corollary is to be understood regarding those parallelograms about the diameter of a square that share a common angle with the whole square, such as the aforementioned parallelograms CI and FH; for the former shares the angle ABD with the square, while the latter shares the angle AED.

Nevertheless, the same is no less true for any parallelograms about the diameter, even when extended, even though they do not share any angle with the square, provided that their sides are parallel to the sides of the square.

For about the diameter AC of the square BD, let there be a parallelogram FH, whether inside the square or outside, which nevertheless has sides parallel to the sides of the square BD. I say that FH is a square.

Since AB and EF are parallel, the angles BAC and FEG will be equal (as corresponding interior and exterior angles); and by the same reasoning, the angles BCA and FGE will be equal. But the angles BAC and BCA are each half of a right angle ($45^\circ$), as has already been shown. Therefore, the angles FEG and FGE will also be half of a right angle, and because of this, the sides EF and FG opposite to them are equal, and the angle F is a right angle.

Therefore, since the sides EF and FG are equal, and are equal to their opposite sides GH and HE, FH will be a square.

Which is what was proposed.

Scholium

This fourth theorem is applied to numbers as follows.

Suppose the number is 10, divided into 7 and 3.

You see, therefore, that the square of the whole number, 100, is equal to the squares of the parts, 49 and 9, together with twice the number 21, which is produced from multiplying 7 by 3.

For 49 + 9 + 21 + 21 = 100.

From this theorem it is very easy to prove that if one straight line is double another straight line, then the square described on the former is four times the square described on the latter.

Let the straight line AB be double the straight line K.

I say that the square on AB is four times the square on K.

For let AB be bisected at C, and let the construction be made as in the theorem.

Then the four parallelograms AG, CI, IF, and FH are all squares and are equal to one another, since all their sides are equal, as may easily be proved from Proposition 34 of Book I, and all their angles are right angles.

Therefore, since the square AD is equal to the four squares AG, CI, IF, and FH, the square on AB is four times the square on AC, that is, on line K, since K is equal to AC. For AC is half of AB

Briefly

From this proposition it follows that the square on AB is equal to the squares on AC and CB, together with twice the rectangle contained by AC and CB.

Now, if the squares on the equal lines AC and CB are equal, and the rectangle contained by the equal lines AC and CB is itself a square and equal to the square on AC, then it follows that the square on AB is four times the square on AC, since it is equal to four squares each equal to the square on AC.

Conversely, we shall show that if one square is four times another square, then the side of the former is double the side of the latter.**

Let the square on AB be four times the square on K.

I say that AB is double K.

For let AB be bisected at C. Then AB is double AC.

Hence, as has just been proved, the square on AB is four times the square on AC.

But by hypothesis it is also four times the square on K.

Therefore the squares on K and AC are equal.

Hence the lines K and AC themselves are equal.

But by construction AB is double AC.

Therefore AB is also double K.

We shall, however, prove all these things by another method in Proposition 20 of Book VI.

Image of the Pages from Clavius for Proposition 4, Book II