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= AB = BC—AČ. 188. Corollary II. Let ABCD (fig. 118) be a square, AC its Fig. 118. diagonal; the triangle ABC being right-angled and isosceles, we 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. 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 III. It has been demonstrated, that the square AH (fig. 109) is equivalent to the rectangle BDEF; now, on Fig. 109 account of the common altitude BF, the square BCGF is to the rectangle BDEF as the base BC is to the base BD; therefore BC: AB:: BC: BD, or, 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 BC AC BC: CD. 190. Corollary IV. 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, AB: AC:: BD: DC, 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. Fig. 110. 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 2 AB = AC + BC -2BC × CD. Demonstration. The proposition admits of two cases. 1. If the perpendicular fall within the triangle ABC, we shall have BD=BC — CD; and, consequently, (182) adding AĎ -2 BD= BC+ CD-2BC x CD; to each member, we have AB + BB = BC + CB+ AB—2BC × CD; but the right-angled triangles ABD, ADC, give AĎ + BĎ=AB, CĎ AB = BC + AC — 2BC × CD. 2. If the perpendicular AD fall without the triangle ABC, we shall have BD = CD-BC; and, consequently, (182) BB = CB+ add to each AĎ, and we shall obtain, as before, AB = BC + AC — 2BC × CD. Fig. 111 THEOREM. 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 (60); 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 AB=BC + AC + 2BC × CD. 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 ver- Fig. 112. tex 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 AB = AE + EB + 2EB × ED; therefore, by adding the corresponding members, and observing -that EB EC, we shall have = AB + AC=2AE + 2EB. 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 hisect each Fig. 113. other in the point E (88), and the triangle ABC gives AB+BC = 2AE + 2BE; the triangle ADC gives likewise AB + DC = 2AB+2DÊ; adding the corresponding members, and observing that BE=DE, we have 4A AB+ AD + DC + BC = 4AE+4DĒ. But 4ÃỄ is the square of 2AE or of AC; and 4DE is the square of BD; therefore the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. Fig 114. THEOREM. 196. The line DE (fig. 114), drawn parallel to the base of a triangle ABC, divides the sides AB, AC, proportionally; so that AD: DB:: AE: EC. Demonstration. Join BE and DC; the two triangles BDE, DEC, have the same base DE; they have also the same altitude, since the vertices B and C are situated in a parallel to the base; therefore the triangles are equivalent (170). The triangles ADE, BDE, of which the common vertex is E, have the same altitude, and are to each other as their bases AD, DB (177); thus ADE: BDE:: AD: DB. The triangles ADE, DEC, of which the common vertex is D, have also the same altitude, and are to each other as their bases AE, EC; that is, ADE: DEC:: AE: EC. But it has been shown, that the triangle BDE = DEC; therefore, on account of the common ratio in the two proportions (111), AD: DB:: AE: EC. 197. Corollary 1. We obtain from the above theorem, by composition (Iv), Fig. 115. (fig. 115), parallels AC, EF, GH, BD, &c., be drawn, these two straight lines will be cut proportionally, and we shall have, For, let O be the point of meeting of the straight lines, AB, CD; in the triangle OEF, the line AC being drawn parallel to the base EF, OE: AE:: OF: CF, or OE: OF:: AE: CF. In the triangle OGH we have likewise OE : EG :: OF : FH, or OE: OF :: EG: FH; therefore, on account of the common ratio OE: OF, these two proportions give ᏁᎬ : CF :: EG: FH. It may be demonstrated, in the same manner, that EG: FH:: GB: HD, and so on; therefore the lines AB, CD, are cut proportionally by the parallels EF, GH, &c. THEOREM. 199. Reciprocally, if the sides AB, AC (fig. 116), are cut pro- Fig. 116 portionally by the line DE, so that AD: DB:: AE EC, the line DE will be parallel to the base BC. Demonstration. If DE is not parallel to BC, let us suppose that DO is parallel to it; then, according to the preceding theoAD: DB :: AO : OC. rem, which is impossible, since of the antecedents AE is greater than AO, and of the consequents EC is less than OC; hence the line, drawn through the point D parallel to BC, does not differ from DE; therefore DE is this line. 200. Scholium. The same conclusion might be deduced from the proportion AB: AD :: AC: AE. For this proportion would give (IV) AB—AD : AD::AC—AE : AE, or BD : AD :: EC: AE THEOREM. 201. The line AD (fig. 117), which bisects the angle BAC of a Fig. 17. triangle, divides the base BC into two segments BD, DC, proportional to the adjacent sides AB, AC; so that, BD : DC :: AB : AC. Demonstration. Through the point C draw CE parallel to AD to meet BA produced. In the triangle BCE, the line AD being parallel to the base (196), GE, BD: DC: AB: AE. |
