The square ACDE is divided into four parts; the first ABIF is the square described upon AB, since AF was taken equal to AB; the second IGDH is the square described upon BC; for, since ACAE, and AB AF, the difference AC-ABAE - AF, which gives BC = EF; but, on account of the parallels, IG = BC, and DG EF, therefore HIGD is equal to the square described upon BC. These two parts being taken from the whole square, there remain the two rectangles BCGI, EFIH, which have each for their measure AB × BC; therefore the square described upon AC, &c. 181. Scholium. This proposition corresponds to that given in algebra for the formation of the square of a binomial, which is thus expressed, (a + b)2 = a2 +2ab+b2. THEOREM. 182. If the line AC (fig. 107) is the difference of two lines AB, Fig. 107. BC, the square described upon AC will contain the square of AB, plus the square of BC, minus twice the rectangle contained by AB and BC; that is, AC or (AB - BC)= AB+ BC-2 AB × BC. Demonstration. Construct the square ABIF, take AE = AC, draw CG parallel to BI, HK parallel to AB, and finish the square EFLK. The two rectangles CBIG, GLKD, have each for their measure AB × BC; if we subtract them from the whole figure --2 ABILKEA, which has for its value AB + BC it is evident, that there will remain the square ACDE; therefore, if the line AC, &c. 183. Scholium. This proposition answers to the algebraic formula (a - b)2=a 2 + b2 — 2 a b. THEOREM. 184. The rectangle contained by the sum and difference of two lines is equal to the difference of their squares; that is, (AB + BC) × (AB — BC) = AB — BC (fig. 108). Demonstration. Construct upon AB and AC the squares ABIF, ACDE; produce AB making BK = BC, and complete the rectangle AKLE. Fig. 108. The base AK of the rectangle is the sum of the two lines AB, BC, its altitude AE is the difference of these lines; therefore the rectangle AKLE = (AB+BC) × (AB — BC). But this same rectangle is composed of two parts ABHE + BHLK, and the part BHLK is equal to the rectangle EDGF, for BH = DE, and BK = EF; consequently AKLE=ABHE+EDGF. Now these two parts form the square ABIF, minus the square DHIG which is the square described upon BC; therefore -2 -2 (AB+BC) × (AB—BC)= AB — BC. 185. Scholium. This proposition agrees with the algebraic formula (a + b) x (a—b) = (a2— b2) (Alg. 34). ig. 109. THEOREM. 186. The square described upon the hypothenuse of a right-angled triangle is equal to the sum of the squares described upon the two other sides. Demonstration. Let ABC (fig. 109) be a triangle right-angled at A. Having constructed squares upon the three sides, let fall, from the right angle upon the hypothenuse, the perpendicular AD, which produce to E, and draw the diagonals AF, CH. The angle ABF is composed of the angle ABC plus the right angle CBF; and the angle HBC is composed of the same angle ABC plus the right angle ABH; hence the angle ABF = HBC. But AB = BH, being sides of the same square; and BF = BC, for the same reason; consequently the triangles ABF, HBC, have two sides and the included angle of the one respectively equal to two sides and the included angle of the other; they are therefore equal (36). The triangle ABF is half of the rectangle BEt, which has the same base BF and the same altitude BD (169). Also the triangle HBC is half of the square AH; for, the angle BAC being a right angle as well as BAL, AC and AL are in the same straight line parallel to HB; hence the triangle HBC and the square AH have the same base BH, and the same altitude AB; therefore the triangle is half of the square. † An abridged expression for BDEF. It has already been proved, that the triangle ABF is equal to the triangle HBC; consequently the rectangle BDEF, double of the triangle ABF, is equivalent to the square AH, double of the triangle HBC. It may be demonstrated, in the same manner, that the rectangle CDEG is equivalent to the square AI; but the two rectangles BDEF, CDEG, taken together, make the square BCGF; therefore the square BCGF, described upon the hypothenuse, is equal to the sum of the squares ABHL, ACIK, described upon the two other sides; or, BC = AB+ AC. 187. Corollary 1. The square of one of the sides of a rightangled triangle is equal to the square of the hypothenuse minus the square of the other side; or AB=BC- -AC. -2 -2 -2 188. Corollary 11. Let ABCD (fig. 118) be a square, AC its Fig. 118. diagonal; the triangle ABC being right-angled and isosceles, we -2 -2 have AC = AB+BC-2AB; therefore the square described upon the diagonal AC is double of the square described upon the side AB. This property may be rendered sensible by drawing, through the points A and C, parallels to BD, and through the points B and D, parallels to AC; a new square EFGH is thus formed which is the square of AC. It is manifest that EFGH contains eight triangles, each of which is equal to ABE, and that ABCD contains four of them; therefore the square EFGH is double of ABCD. --2 --2 Since AC: AB::2:1, we have, by extracting the square root, AC: AB::2:1; therefore the diagonal of a square is incommensurable with its side (Alg. 99). This will be more fully developed hereafter. 189. Corollary . It has been demonstrated, that the square AH (fig. 109) is equivalent to the rectangle BDEF; now, on Fig. 105 account of the common altitude BF, the square BCGF is to the rectangle BDEF as the base BC is to the base BD; therefore or, --2 the square of the hypothenuse is to the square of one of the sides of the right angle, as the hypothenuse is to the segment adjacent to this side. We give the name of segment to that part of the hypothenuse cut off by the perpendicular let fall from the right angle; thus BD is the segment adjacent to the side AB, and DC the segment adjacent to the side AC. We have likewise Fig. 110. Fig. 111. 190. Corollary iv. -2 -2 BC: AC:: BC: CD. The rectangles BDEF, DCGE, having also the same altitude DE, are to each other as their bases BD, CD. Now these rectangles are equivalent to the squares AH, AI, therefore," or, the squares of the two sides of a right angle are to each other, as the segments of the hypothenuse adjacent to these sides. THEOREM. 191. In a triangle ABC (fig. 110), if the angle C be acute, the square of the side opposite to it will be less than the sum of the squares of the sides containing it, and, AD being drawn perpendicular to BC, the difference will be equal to double the rectangle BC x CD, or, the perpendicular fall within the triangle ABC, we shall have BD = BC — CD, and consequently (182) -2 -2 BD2= BC + CD — 2BC × CD ; adding AD to each member, we have -2 -2 AD + BD = BC + CD + AD-2BC x CD ; -2 but the right-angled triangles ABD, ADC, give AD+BD=AB, CD+AD=AC; therefore -2 AB = BC + AC — 2BC × CD. 2. If the perpendicular AD fall without the triangle ABC, we shall have BD CD- BC, and consequently (182) add to each AD, and we shall obtain, as before, 192. In a triangle ABC (fig. 111), if the angle C be obtuse, the square of the side opposite to it will be greater than the sum of the squares of the sides containing it, and, AD being drawn perpendicular to BC produced, the difference will be equal to double the rectangle BC × CD, or, AB=AC+BC+2BC × CD. Demonstration. The perpendicular cannot fall within the triangle; for if it should fall, for example, upon E, the triangle ACE would have at the same time a right angle E and an obtuse angle C, which is impossible (75); consequently it falls without, and we have BD = BC + CD, and from this (180) Adding to each term AD, and making the reductions as in the preceding theorem, we obtain 193. Scholium. The right-angled triangle is the only one in which the sum of the squares of two of the sides is equal to the square of the third; for, if the angle contained by their sides be acute, the sum of their squares will be greater than the square of the side opposite; if it be obtuse, the reverse will be true. THEOREM. 194. In any triangle ABC (fig. 112), if we draw from the vertex Fig. 112. to the middle of the base the line AE, we shall have Demonstration. Let fall the perpendicular AD upon the base BC, the triangle AEC will give (191), --2 AC = AE+EC-2EC × ED; the triangle ABE will give (192), therefore, by adding the corresponding members, and observing that EB EC, we shall have H 195. Corollary. In every parallelogram the sum of the squares of the sides is equal to the sum of the squares of the diagonals. For the diagonals AC, BD (fig. 113), mutually bisect each Fig. 113, other in the point E (88), and the triangle ABC gives |