The angles at the base of an isosceles triangle are equal
This is the first "real" theorem in the Elements. The resemblance of its figure to a bridge, plus its difficulty for beginners, led to its being called the pons asinorum, or "bridge of asses", a kind of initiation ceremony (or so I am told; the same epithet has been applied to other propositions). The first three propositions of Book I are problems, proving the ability to cut off from any line a length equal to a shorter line. Then, proposition 4 states that if two triangles have an angle equal, and the sides embracing this angle equal to each other, each to each respectively, then the triangles are equal, and the other parts are also equal. In modern school geometry, the term congruent is for some reason introduced to describe the relation of the two triangles, so that a complicated word can serve for a simple one. This theorem can be referred to as SAS, side-angle-side, and is one of the most common ways to prove two triangles equal. Euclid's proof of it is by superposition (one must start somewhere), which some commentators feel is cheating. Of course, it is not, since this is an old and time-honoured method of proof that is avoided in Euclid as far as possible.
Now let's get down to the proposition. The Greek is at the right, and you can see it is not trivial. However, we shall go through it word by word, since it is an excellent example of a proposition of the Elements, with all the usual parts included. We have to prove that the angles at the base of an isosceles triangle (one with two equal sides; the base is the third side) are equal. In pre-Euclidean geometry, this was probably proved by superposition. That is, the triangle was imagined to be flipped over and then shown that it would coincide with its earlier location. Euclid has just made a proof in this way to get SAS, and now shows that he has a powerful tool that can be used in place of superposition.
The statement of the theorem is: "of isosceles triangles, the angles at the base are equal to one another, and the equal lines having been extended, the angles under the base will be equal to one another." This is rather literal, but the English words have been put in a better order. You should have no trouble with the vocabulary, except maybe for the participle of prosekba/llw, to extend or produce. Note that it occurs in a genitive absolute.
The next paragraph applies the theorem to the figure. "Let there be an isosceles triangle ABΓ having the AB side equal to the AΓ side, and let the lines be extended on straight lines to AB, AΓ the lines BΔ, ΓGE. I say, that on one hand the angle ABΓ is equal to the angle AΓB, and the ΓBA to the BΓE." Once again, this is not quite literal, but you should be able to figure it out. What allows us to extend the sides in a straight line?.
Now we have the construction. Of course, we have already extended the sides, which is a construction, but this was necessary in order to state the theorem. We move on to the two remaining lines. "Choose, then, on BΔ by chance a point Z, and cut off on the greater AE to the lesser AZ [the line] AH, and join ZΓ, HB." We have three aorist third-person imperatives in this paragraph, all of which we have studied before. We have lesser and greater, two much-used terms. The word tucon is actually an adverb, meaning by chance, randomly, where we would probably use an adjective in English, a "random point".
At last, we have the demonstration or proof. What we are going to do is first prove that triangles ABH and AΓZ are equal, using SAS. Then we will show that triangles BΓH and ΓBZ are equal as well by SAS, using the previous result (that the angles at Z and H are equal). The conclusion for the angles under the base then follows at once. The angles above the base are found by subtracting equals from equals, so these angles are also equal. Think through this before you proceed. Now that we know what we are going to do, we can set it out formally and justify each step.
We begin: "Since, therefore, (me/n) AZ is equal to AH, (de/) AB to AΓ, the two ZA, AΓ are equal to the two HA, AB each to each; And ZAH embraces a common angle; therefore, the base ZΓ is equal to the base HB, and triangle AZΓ is equal to triangle AHB, and the remaining angles to the remaining angles will be equal, each to each, under which the equal sides subtend, (me/n) AΓZ to ABH, (de/) AZΓ to AHB." We have now completed the first step of the proof. Note how me/n and de/ are used to separate and identify corresponding things, where we resort to punctuation or extra words in English.
Continuing: "Also, since the whole AZ is equal to the whole AH, of which AB is equal to AΓ, the remainder, therefore, BZ is equal to the remainder ΓH. And it has been shown ZΓ is equal to HB; thus the two BZ, ZΓ are equal to the two ΓH, HB, each to each; also, the angle BZΓ is equal to angle ΓHB, and the base common of them [is] BΓ; and therefore the triangle BZΓ will be equal to the triangle ΓHB, and the remaining angles will be equal to the remaining angles each to each, which the equal sides subtend;" Now we have finished the second part of the proof we sketched at the beginning. In the phrase "each to each", note that the two words are pronounced alike: hekate'ra hekate' ra, since the iota subscript is silent.
Now for the final push: "Equal therefore is ZBΓ to HΓB, BΓZ to ΓBH. Since, accordingly, the whole angle ABH to the whole angle AΓZ has been proved equal, of them the ΓBH equal to the BΓZ, the remainder, therefore, ABΓ is equal to the remainder AΓB; and they are at the base of the triangle. And ZBΓ has been shown equal to HΓB; and they are under the base." The demonstration is now complete, and we have shown that the pairs of angles are equal. All that remains is to sum up and close.
"Therefore, of isosceles triangles the angles at the base are equal to one another, and having extended the equal lines, the angles under the base will be equal to one another: which was to be shown." The proposition is complete.
Composed by J. B. Calvert
Created 25 September 2000
Last revised 16 June 2002