§ 192. f.) If aBAC = 90°, and AB = = hm AB2? LXXXIV. E = AC, then BC2: Draw CD || AB, and BD || AC. Join AD; and through the points A, D, B, and C, draw EH and FG || BC, B and EF and HG || AD. Point out in the figure the squares of AB and BC, and show from the figure that BC2 = 2AB2. AB2= F What part, then, is any square of the square of its diagonal? How many identical triangles does Fig. LXXXIV. contain? What squares in the figure contain two of these triangles? four? eight? COR. II. Any square is half the square of its diagonal. g.) If one side of a square is 5 ft., what is the square of its diagonal? If one side is 8 ft.? If one side is 10 ft.? What is the base of a square, if the square of its diagonal is 72 sq. yds.? if the square of its diagonal is 98 sq. rods? if the square of its diagonal is 242 sq. rods. What is the distance between the opposite corners of a square field containing 50 sq. rods? of one containing 1 acre and 40 sq. rods? of one containing § of an acre? § 193. h.) In Fig. LXXXIX., AB2 And And BCX BL? 187. AC2 BC X LC? BL2 + LC2 + 2AL?? 183. BC2 BL2+ LC2 + 2BL × LC? 180. 20. f,k. COR. III. If the hypotenuse of a right triangle be divided into two segments by a perpendicular drawn from the right angle, (1.) the square of either of the other sides is equal to the rectangle of the hypotenuse into its segment adjoining (e) Lat. segmentum, portion, division, part cut off, from seco, to cut. that side; and (2.) the square of this perpendicular is equal to the rectangle of the two segments. Whatever part BL is of BC, is AB2 the same part of BC2? Show from Cor. III. 2., that, if BLLC, then AL= BL Compare § 131. § 194. k.) The following is Euclid's Method of demonstrating Theor. III. Draw AN || BD; and join AD, FC, AE, and HB. Show that tABD and tFBC are identical; and that, therefore (since tABD = BN, and F G XC. H Ꭰ N E Show, in like manner, that tACE 195. 1.) Third Method. Join OP, in Fig. LXXXIX., and show that BOPC is the square of BC. Show that the triangles BFO, OKP, PHC, BAC, AGK, and KJA are identical; and that, if you take the first three from the pentagon BFKHC, BC2 remains, while, if you take from it the last three, AB2 and AC2 remain. ·. BC2= AB2+AC2. .. § 196. m.) Fourth Method. Upon the hypotenuse BC, describe the square BDEC; and through D and E, draw FG and GH parallel to AC and AB, and forming, with these lines produced, the quadrilateral AFGH. Show that the triangles ABC, FDB, GED, and HCE are identical, and are together 2AB X AC; so that AFGH is = BC2 B A LXXXVIII. 20. f. +2ABX AC. Show that AFGH is the square of AB + AC, and is, therefore, = AB2 + AC2 + 2AB X AC. ... BC2 = AB2 + AC2. § 197. n.) Fifth Method. Divide BCDE, the square of the hypotenuse BC, as in Fig. xci. (where the angles DEH, CDI, and BCI are each CBA). Put tIDC into the position ABC, and tHED into the position FEB; and show that you have thus E 20. f. XCI. D H C made, from the square of BC, the two squares of AC and AB. Illustrate this method by the actual dissection of a square drawn upon paper. § 198. o.) This Proposition, which has been named, from its discoverer, the Pythagorean, is the most celebrated in Geometry; and there is none which has more exercised the ingenuity of mathematicians in devising different methods of proof. Hoffmann, in his Treatise upon the Proposition (Mayence, 1819), exhibits no fewer than thirty-two methods differing more or less from each other. Of these, the methods given above are the most celebrated and beautiful. (a+b)2 + (a - b)2 m 2a2 + 262 ? Dividing by 2, ±(a+b)2 +§(a - ·b)2 m a2 + b2? COR. IV. (1.) The squares of two lines are together equal to half the squares of their sum and difference. (2.) The squares of the sum and difference of two lines are together equal to twice the squares of the lines. Apply this Corollary to two lines, one of which is 6 ft. and the other 4 ft. Apply it to two lines, one of which is 9 ft. and the other 3 ft. § 200. q.) Each part of Cor. iv. may be proved by a separate and beautiful application of Theor. III. (1.) Let the two lines be AB and BC, joined at B so as to form one straight line. Then their sum will be AC, and their difference DB, obtained by taking, from AB, ADBC. Bisect AC in E. Is DB also bisected in E? Why? At E raise EF and = AE; and join FA and FC. Draw BG || EF, meeting FC at G; and draw GH || AB. What lines in the figure are each equal to AE? to BC? to EB? What right-angled isosceles triangles are there in the figure? Join AG. What kind of triangles are AFG and ABG? And ... But And But And AB2 + BG2 AG2 = = square of what line? squares of what two other lines? AB2 + BC2 (= BG2) ≈≈≈ AF2 + FG2 ? AF2 twice the square of what line? FG2= twice the square of what line? AB2 + BC2 ≈ 2AE2 + 2EB?? 2AE2= half of what square? 2EB2 half of what square? = AB2BC2 ¿AC2 + ¿DB?? (2.) Let the two lines be AB and BC, joined at B so as to form one straight line. Then their sum will be AC, and their difference DC, obtained by taking, from BC, BD = AB. 192.j. 179. b. At B raise BE and = AB. F Join AE, and through D draw EDF meeting at F with the line GCF drawn AC through pC. Draw EG || BC, and join AF. = Show, by a method of proof similar to the preceding, that AC2 + DC2 (= CF2) are = AF2; and, therefore, AE2+ EF2; and, therefore, 2AB2 + 2BC2. Three From aA, either of the angles adjacent to AB, draw AF the opposite side BC, produced if necessary. cases will be observed. (1.) If aC is obtuse, AF falls without the triangle on the side of C. (2.) If aC and aB are both acute, AF falls within the triangle. (3.) If aB is obtuse, AF falls without the triangle on the side of AB. The mode of investigating the three cases is the same, and the result the same except in the sign of one of the terms. As aAFB = 90°, AB2 AF2+ BF2 ? But, in Case 1, as BF: CF+ BC, But BF2 CF2+ BC2 + 2BCX CF? AB2 AF2+ CF2+ BC2+2BCX CF? 188. 180. 18. a. 188. AB2 AF2+ CF2+ BC2 — 2BC X CF? AB2 AC2+ BC2 — 2BC × CF? 183. How does the value of AB2 in Cases 2 and 3 differ from its value in Case 1? Universally, AB2 = AC2 + BC2 + or 2BC X CF. Trace upon the figure, and write out separately, the demonstration of each case. |