This page will hightlight some propositions from Book I. This is a work in progress and may grow over time.
Proposition 35 is a theorem about the areas of two parallelograms.. The figure below shows two parallelograms, ABCD and EBCF. Both parallelograms share the same base BC. This also means that AD and EF are also the same. The theorem considers what will happen to the area of one of the parallelograms if it leans to the right, with the constraint that the height of the parallelogram stays the same. Does the area get smaller, larger, or does it stay the same as the parallelogram leans over?
Here is a more extreme case of the parallelogram leaning over from which it might no be so obvious as to what is happening to the area.
If we were to start with two squares with the same area and then lean one of the squares to the right, would its area change, assuming we keep the height the same?
For those who remember their High school geometry may remember that the area of a parallelogram is the base times the height of the parallelogram. You'll notice that the height doesn't change during the shear and we've fixed the base, hence the area cannot change.
Proclus (412–485 C.E) wrote an extensive commentary on Euclid and in one section looked at proposition 35 in detail. He recognized a possible confusion (which we still see in High School students today) as to why the area should remain the same even though the shape and side lengths change. On first glance, this is quite counterintuitive, especially in the extreme example shown above. In fact, in the limit as we shear more and more, the apparent distance between the two vertical lines tends to zero, as point F extends out to infinity.
The proof that Euclid gave is interesting.
The first thing to note is that the opposite sides are equal, that this AD = BC and EF = BC. This also means that AD = EF.
Let us draw a line between D and E. The segment DE is part of AE and DF, that is:
AE = AD + DE and DF = DE + EF
therefore AE = DF. We also note that AB = DC because these are opposite sides of a parallelogram ABCD.
If you look carefully, you can see two triangles, these are EAB and FDC. The angles EAB and FDC are equal to each other because AF and BC are parallel and AB = DC.
We have two sides and one angle; thus, by the Side-Angle-Side rule, the triangles EAB and FDC are the same. Therefore, their areas are the same.
If we remove the area corresponding to the small triangle DEG, the area of what remains of the two triangles must still be equal. That is, the areas ABGD and EGCF must be equal to each other. Finally, if we add the area of the lower triangle GBC to ABGD and EGCF, we are adding the same amount of area to both sections. However, adding GBC to ABGD and EGCF gives us back the two parallelograms, which must therefore have the same areas.
Hence, the two parallelograms have the same area.
Here is another way, and possibly a simpler way to explain the proof using algebra. The letters in the figure below represent areas. For example, 'a' is the area of the blue portion.
The red and blue lines are equal since they are opposite sides of a parallelogram. In addition, the purple and orange lines marking the length of the tops of the two triangles are also equal in length.
The two marked angles are also congruent. These observations let us state that the two triangles a+c and b + c are congruent using SAS. Hence, we can write:
a + c = b + c
or
a = b, that is, the blue and pink areas have the same area.
Now add area d to both sides:
a + d = b + d
but a + d is the area of the left-hand parallelogram, and b + d is the area of the right-hand parallelogram. Therefore, both parallelograms are equal to each other.
Sometime in the 20th century this proposition was dropped from geometry books. Instead, it was replaced, at least in High School math, by the simple formula that the area of a parallelogram is the base times the height. Since the base doesn't change and the height is constant no matter what the shear, the area stays constant.
The figure below shows the various ways in which proposition 35 has been drawn in a variety of manuscripts dating back to around 450 AD.. The figure below shows diagrams from Greek manuscripts plus one Latin manuscript. These are listed in order of date. The first is from Proculus' commentary of Euclid's Elements. from about 450 AD. We're assuming here that the figure has been preserved as it was originally drawn in 450 AD. Notice that the left most parallelogram a rectangle. When we jump to the 888AD manuscript, we see that the parallelogram has changed to a square. This style continues for the next 700 years or so. There are, however, three interesting differences. The first is that Billingsley's English version has a similar style to the one used by Proclus. As far as I can tell, this style is never used again.
The second interesting change is found in the 12th century Greek to Latin translation. The next time we see it is in Commandino's 1572 edition. Does this suggest that Commandino used the 12th century Latin edition? We see it next in Simson's 1756 edition, and after that, this is the only style we appear to see in textbooks.
One wonders why the change occurred in the 12th-century Greek to Latin translation. It is known that whoever did the translation appeared to be knowledgeable in mathematics. It may be that the translator thought that the new style was better.
In the next section, we'll look at the Arabic-derived manuscripts.
I noticed an interesting difference with Arabic and Arabic-derived manuscripts. All these manuscripts use either the Proculus or the square style. But one noticeable difference is that the orientation is the mirror of the Greek texts. This even extends to the Latin texts that were derived from Arabic sources. This particular style could be used perhaps as a diagnostic of Arabic derived texts.
An examination of other figures from Book I appears to reveal mirror images in the Arabic or Arabic-derived texts.