Plane Geometry - Bruce E. Shapiro

196 SECTION 38. THE PYTHAGOREAN THEOREMEuclid, of course, did not state the Pythagorean theorem in terms of thesum of the squares of the edges; to do so would have required algebra,which was not invented for another thousand years after Euclid. Instead,the theorem was expressed in terms of area.Theorem 38.9 (Euclid’s Version of the Pythagorean Theorem) Thearea of the square on the hypoteneuse of a right triangle is equal to the sumof areas of the squares on the legs.Proof. (Second Proof of the Pythago- rean Theorem, due to Euclid6 )Let △ABC be a right triangle with right angle at C. We construct thesquares on each edge and define the points as indicated in figure 38.5.Point J is the foot of the perpendicular to AB dropped from C, and pointK is the intersection of the same perpendicular with DE.From the Euclidean Area Postulate,α(□ABED) = α(□AJKD) + α(□JBEK) (38.1)Since ∠ACB is a right angle and □ACHI is a square, H ∈ ←→ BC and AI ‖←→BC.Thus, since triangles △IAC and △IAB each have base IA and height AC,they both have the same area:Furthermore, since IA = AC, AB = AD, andthen by SASα(△IAC) = α(△IAB) (38.2)∠IAB = 90 + ∠CAB = ∠CAD△IAB ∼ = △CADand since congruent triangles have equal areas (by the neutral area postulate),α(△IAB) = α(△CAD) (38.3)Since △CAD and △JAD each have base AD and height AJ, they haveequal area,α(△CAD) = α(△JAD) (38.4)Combining equations 38.2, 38.3 and 38.4 givesα(△IAC) = α(△JAD) (38.5)6 Version given in Book 1 of the Elements; another proof due to Euclid is in Book VI.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012