Well, we've reached the end of Book I of the Elements. We haven't covered every proof and construction in Book I, of course, but we've talked about 12 of them specifically, mostly from Book I, and have used or given honorable mention to 8 more.
The last two theorems at the end of Book I involve proofs of the well known Pythagorean Theorem (I.47) and its converse (I.48). But there is a more general theorem and proof from Pappus given in the notes of Heath's translation of the Elements, which shows that given any triangle with parallelograms constructed on two of its sides, one can identify a parallelogram constructed on the third side which has an area equal to the sum of the areas of the other two parallelograms. The Pythagorean Theorem can be seen, then, as a special case of this more general theorem.
Pappus Shows Up the Goofy Pythagoreans
We start with any old triangle, \(\small\Delta\mathtt{ABC}\), and build any old parallelograms on any two of its sides—in this case, we build parallelogram \(\small\mathtt{ADEB}\) on \(\small\overline{\mathtt{AB}}\) and parallelogram \(\small\mathtt{CFGB}\) on \(\small\overline{\mathtt {BC}}\).
We then extend lines \(\small\mathtt{DE}\) and \(\small\mathtt{FG}\) to intersect at a point \(\small\mathtt{H}\). And then a bunch of other stuff. Draw \(\small\overline{\mathtt{HK}}\) through point \(\small\mathtt{B}\) to intersect with side \(\small\mathtt{AC}\) at \(\small\mathtt{K}\). Finally, construct \(\small\overline{\mathtt{AL}}\) and \(\small\overline{\mathtt{CM}}\) both parallel to \(\small\overline{\mathtt{HK}}\) and join \(\small \mathtt{L}\) and \(\small\mathtt{M}\) to then form quadrilateral \(\small\mathtt{ALMC}\).
We know that \(\small\mathtt{ALHB}\) is a parallelogram, because we constructed its opposite sides to be parallel. So, using Proposition 34 in Book I, we know that \(\small\overline{\mathtt{AL}} \cong \overline{\mathtt{HB}}\). Similarly, \(\small\mathtt{CMHB}\) is a parallelogram, so we know that \(\small\overline{\mathtt{CM}} \cong \overline{\mathtt{HB}}\). This means that \(\small\overline{\mathtt{AL}}\) and \(\small\overline{\mathtt{CM}}\) are both congruent and parallel to each other, since they are both congruent and parallel to \(\small\overline{\mathtt{HB}}\). Proposition 33, which we haven't discussed, tells us that lines connecting congruent parallel lines are themselves congruent and parallel. So, \(\small\overline{\mathtt{LM}} \cong \overline{\mathtt{AC}}\) and \(\small\overline{\mathtt{LM}} \parallel \overline{\mathtt{AC}}\), meaning that \(\small\mathtt{ALMC}\) is a parallelogram.
Okey doke. More stuff we can know. Well, Proposition 36 (another one we haven't gone over) tells us that parallelograms that share a base and are "in the same parallels" have equal areas. So parallelograms \(\small\mathtt{DABE}\) and \(\small\mathtt{LABH}\) have equal areas, since they share \(\small\overline{\mathtt{AB}}\) as a base and are in the parallels \(\small\overleftrightarrow{\mathtt{AB}}\) and \(\small\overleftrightarrow{\mathtt {DH}}\).
Interestingly, if we draw point \(\small\mathtt{N}\) at the intersection of \(\small\overline{\mathtt{HK}}\) and \(\small\overline{\mathtt{LM}}\), we can deduce that parallelograms \(\small\mathtt{LABH}\) and \(\small\mathtt{LAKN}\) are also equal in area, since they share \(\small\overline {\mathtt{LA}}\) as a base and are in the parallels \(\small\overleftrightarrow{\mathtt{LA}}\) and \(\small\overleftrightarrow{\mathtt{HK}}\).
Well, since parallelograms \(\small\mathtt{DABE}\) and \(\small\mathtt{LABH}\) have equal areas, and parallelograms \(\small\mathtt{LABH}\) and \(\small\mathtt{LAKN}\) are also equal in area, we can say that parallelograms \(\small\mathtt{DABE}\) and \(\small\mathtt{LAKN}\) are equal in area.
Pappus, Take Us Home
And this is far more general than what is proved in the Elements about squares in the case of right-angled (triangles). [drops mic]
If you're interested in more extensions of the Pythagorean Theorem, check out the "blob" extension below. And by all means, get your hands on a copy of a translation of Euclid's Elements. It's great reading!
Image credit: Garrett Coakley