1. The Elements, Book I – the “destination” – Proposition 47 MONT 104Q – Mathematical Journeys: Known to Unknown

2. The goal we have worked toward Proposition 47. In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the two other sides. The “Theorem of Pythagoras,” but stated in terms of areas (not as an algebraic identity!) The proof Euclid gives has to rank as one of the masterpieces of all of mathematics, although it is far from the simplest possible proof. The thing that is truly remarkable is the way the proof uses just what has been developed so far in Book I of the Elements.

3. Euclid's proof, construction Let the right triangle be ABC with right angle at A Construct the squares on the three sides (Proposition 46) and draw a line through A parallel to BD (Proposition 31)

4. Euclid's proof, step 1 <BAG + <BAC = 2 right angles, so G,A,C are all on one line (Proposition 14), and that line is parallel to FB (Proposition 28) Consider FBG

5. Euclid's proof, step 2 Proposition 37 implies areas of FBG and FBC are equal

6. Euclid's proof, step 3 AB = FB and BC = BD since ABFG and BCED are squares <ABD = <FBC since each is a right angle plus <ABC Hence BFC and ABD are congruent (Proposition 4 – SAS)

7. Euclid's proof, step 4 Proposition 37 again implies BDA and BDM have the same area. Hence square ABFG and rectangle BLKD have the same area (twice the corresponding triangles – Proposition 41)

8. Euclid's proof, conclusion A similar argument “on the other side” shows ACKH and CLME have the same area Therefore BCDE = ABFG + ACKH. QED