Here we are in book 1 of Euclid’s Elements. Just for fun, I decided to skip ahead a few to the first interesting proof in the Elements, which is Proposition 5 involving Isosceles triangles. I will come back to Proposition 2 through 4 later.

__Proposition 5__

__Proposition 5__

**In isosceles triangles, the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.**

**In isosceles triangles, the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.**

*Let ABC be an isosceles triangle having the side AB equal to the side AC, and let the straight-lines BD and CE have been produced in a straight-line with AB and AC (respectively). I say that the angle ABC is equal to ACB, and (angle) CBD to BCE.*

I find it interesting that Euclid speaks in the first person in the proof. Not that their is anything wrong with that. It just seems informal for someone who is probably writing in the Library of Alexandria, Egypt (Pre-Fire).

I have made markings in green to show the given information and the red marks are the angle which must be proven congruent. (Euclid says equal, but we all know equal means same measure and same location, and we know he meant congruent which means “equal in measure”. Another geometric term which has apparently evolved in 2500 years.)

Another way to word this proof is: The base angles of an Isosceles triangle are congruent.

Back to Euclid:

For let the point F have been taken at random on BD, and let AG have been cut off from the greater AE, equal to the lesser AF. Also, let the straight-lines F C and GB **have been joined. ****In fact, since AF is equal to AG, and AB to AC, the two (straight-lines) FA, AC are equal to the two (straight-lines) GA, AB, respectively. ****They also encom- pass a common angle, F AG. Thus, the base F C is equal to the base GB, and the triangle AF C will be equal to the triangle AGB, and the remaining angles subtendend by the equal sides will be equal to the corresponding remain- ing angles**

It seems his strategy is to show that triangle FBC is congruent to triangle GCB (by side-side-side) and therefore the obtuse angles in the drawing FBC and GCB will be congruent and, by linear pairs, the bas angles of the original triangle will be congruent. One more drawing. Congruent sides are :