(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? AB2 + BG2 square of what line? And ... But And But AG2 = squares of what two other lines? AB2 + BC2 ≈ 2AE2 + 2EB2? . 2AE2 half of what square? 2EB2 = half of what square? AB2+ BC2 ≈ ¿AC2 + ¿DB2? (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. At B raise BE and = AB. Join AE, and through D draw A XCIII. E 192. f. 179. b. 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. From aA, either of the angles adjacent to AB, draw AF the opposite side BC, produced if necessary. Three 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. BF2 CF2+ BC2 + 2BC × CF ? AB2 AF2 + CF2 + BC2 + 2BC× CF? 188. 180. 18. a. 188. AB2 AF2+ CF2+ BC2 2BC X CF? AB2 AC2+ BC2-2BC x CF? 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. § 202. THEOR. IV. The square of any side of a triangle subtending an OBTUSE angle is GREATER, and of any side subtending an ACUTE angle LESS, than the sum of the squares of the other two sides, by twice the rectangle of either of these sides into the distance between a perpendicular let fall upon it from the opposite vertex and the given angle. [Proved by means of Theor. III., and $$ 180, 183.] § 203. a.) 1. Given, in tABC (Fig. xciv.), AB2 > AC2 +BC2. 2. Show by the same mode of reasoning, that, if AB2 < AC2+ BC, then aC < 90°; and, if AB2 = AC2 + BC2, then aC = 90°. Hence (uniting §§ 188, 202, 203. a), b.) COR. 1. The square of any side of a triangle is =, >, or the sum of the squares of the other two sides, according as the angle which it subtends is > or < a right angle; and the converse. = c.) Show, from § 78, that the sides containing the right angle AFB are greater than those containing the obtuse angle ACB, and less than those containing the acute angle AcB. or § 204. d.) If the formula above (AB2 = AC2 + BC2+ 2BC × CF) were applied to the hypotenuse of a right triangle, as the perpendicular AF must now fall upon AC, how much would CF be? How much, then, would + or 2BC × CF be? Do you obtain, then, the same result as in § 187? This formula, therefore, may be applied to any side of any triangle. § 205. e.) As, in Fig. xciv., AB2 = AF2 + BF", AF2 + CF2, AC2 = and 20. f. COR. II. The difference between the squares of any two sides of a triangle is the same with the difference between the squares of their distances from a perpendicular let fall upon the third side from the opposite angle. PROPOSITION V. $206. Given, any triangle ABC, with a line drawn from the middle of any side BC to the opposite angle. B D C B DFC Required, the sum of the squares of AB and AC, in terms of BC and AD. If AD is BC, as in No. 1, then AB2 what two squares? And AC2 what two squares? = AB2 + AC2=what four squares? 188. If AD is not = = BC, as in No. 2, draw AF BC. 202. Then AB2 what two squares + what rectangle? And AC what two squares In adding these equations, what terms cancel each other? 21. a. But, in the four squares, in both cases, what square is taken twice? And BD2 + DC2 ≈≈ 2BD2? AB2 + AC2 ≈ 2BD2 + 2AD2? § 207. THEOR. V. The squares of any two sides of a triangle are together equal to twice the squares N of half the third side, and of the straight line drawn from the middle of this side to the opposite angle. [Proved by applying Theorems III. and IV.] § 208. a.) What kind of a triangle is ABC in Fig. xcv., No. 1? If AB and AC are each 91, 205. 5 ft. and BC = 6 ft., what is the length of AD? What is the area of the triangle? 173. If BC = 12 ft. and AD equal sides AB and AC? 8 ft., what is the length of the There is a field in the shape of an isosceles triangle, of which the area is 7 acres and the altitude 40 rods; what is the length of each side? If the three sides of an isosceles triangle are given, how do you find the altitude? the area? b.) If, in No. 2, AB 9 ft., AC7 ft., and AD2: 29 ft., what is the length of BC? = c.) If aBAC = 90°, then 2BD2 + 2AD2 = what single square? § 209. d.) Given, any parallelogram ABCD. Required, the sum of the squares of its sides, in terms of the diagonals. 131, 179. XLIV. A B As, in tABC, BF bisects AC, AB2 + BC2 So, in tCDA, CD2 + DA2 = what? D C .. Adding the equations, AB2 + BC2 + CD2 + DA2 m 4AF2+2BF2 + 2FD2? AB2+ BC2+CD2 + DA2 ≈ AC2 + BD2? 179. COR. In any parallelogram, the squares of the sides are together equal to the squares of the diagonals. If AB: = 5 ft., BC = 5 ft., and BD = 6 ft., what is the = length of AC? If AD = 6 ft., DC 8 ft., and BD2 = 79 sq. ft., what is the length of AC? |