Fig 119. = But the triangle ACE is isosceles; for, on account of the parallels AD, CE, the angle ACE DAC, and the angle AEC=BAD (76); and, by hypothesis, DAC = BAD; therefore the angle ACE= AEC, and, consequently, AE =AC (48); substituting, then, AC for AE in the preceding proportion, we have BD:DC::AB: AC. THEOREM. 202. Two equiangular triangles have their homologous sides proportional, and are similar. Demonstration. Let ABC, CDE (fig. 119), be two triangles, which have their angles equal, each to each, namely, BAC-CDE, ABCDCE, and ACB-DEC; the homologous sides, or those adjacent to the equal angles, will be proportional; that is, BC: CE:: BA: CD :: AC: DE. Let the homologous sides BC, CE, be in the same straight line, and produce the sides BA, ED, till they meet in F. Since BCE is a straight line, and the angle BCA — CED, it follows that AC is parallel to DE (76). Also, since the angle ABC= DCE, the line AB is parallel to DC; therefore the figure ACDF is a parallelogram. In the triangle BFE, the line AC being parallel to the base FE, BC: CE :: BA : AF (196); substituting in the place of AF its equal CD, we have BC: CE: BA: CD. In the same triangle BFE, BF being considered as the base, since CD is parallel to BF, BC: CE :: FD: DE. Substituting for FD its equal AC, we have From these two BC: CE, we have BC: CE:: AC: DE. proportions, which contain the same ratio AC: DE:: BA: CD. Hence the equiangular triangles BAC, CDE, have the homologous sides proportional. But two figures are similar, when they have, at the same time, their angles equal, each to each, and the homologous sides proportional (162); therefore the equiangular triangles BAC, CDE, are two similar figures. 203. Corollary. In order to be similar, it is sufficient that two triangles have two angles of the one respectively equal to two angles of the other; for then the third angles will be equal, and the two triangles will be equiangular. 204. Scholium. It may be remarked, that in similar triangles the homologous sides are opposite to equal angles; thus, the angle ACB being equal to DEC, the side AB is homologous to DC; likewise AC, DE, are homologous, being opposite to the equal angles ABC, DCE. Knowing the homologous sides, we readily form the proportions; AB: DC :: AC: DE ::BC : CE, THEOREM. 205. Two triangles, which have their homologous sides proportional, are equiangular and similar. Demonstration. Let us suppose that BC: EF :: AB : DE :: AC : DF (fig. 120); the triangles ABC, DEF, will have their angles equal, namely,. A=D, BE, C=F. Make, at the point E, the angle FEG = B, and at the point F, the angle EFG = C; the third angle G will be equal to the third angle A, and the two triangles ABC, EFG, will be equiangular; whence, by the preceding theorem, BC: EF:: AB: EG; but, by hypothesis, BC : EF :: AB: DE; consequently EG = DE. We have, moreover, by the same theorem, BC: EF :: AC: FG ; but, by hypothesis, BC: EF:: AC: DF; consequently FG=DF; hence the triangles EGF, DEF, have the three sides of the one equal to the three sides of the other, each to each; they are therefore equal (43). But, by construction, the triangle EGF is equiangular with the triangle ABC; therefore the triangles DEF, ABC, are, in like manner, equiangular and similar. 206. Scholium. It will be perceived, by the two last propositions, that, when the angles of one triangle are respectively equal to those of another, the sides of the former are proportional to those of the latter, and the reverse; so that one of these conditions is sufficient to establish the similitude of triangles. This is not true of figures having more than three sides; for, with respect to those of only four sides, or quadrilaterals, we may alter the proportion of the sides without changing the angles, or change the angles without altering the sides; thus, from the angles being equal, it does not follow that the sides are proportional, or the reverse. Fig. 120. We see, for example, that, by drawing EF (fig. 121) Fig. 21. parallel to BC, the angles of the quadrilateral AEFD are equal Fig. 122. to those of the quadrilateral ABCD; but the proportion of the sides is different. Also, without changing the four sides AB, BC CD, AD, we can bring the points B and D nearer together, or remove them farther apart, which would alter the angles. 207. Scholium. The two preceding theorems (202, 205), which, properly speaking, make only one, added to that of the square of the hypothenuse (186), are, of all the propositions of geometry, the most remarkable for their importance, and the number of results that are derived from them; they are almost sufficient, of themselves, for all applications, and for the resolution of all problems; the reason is, that all figures may be resolved into triangles, and any triangle whatever into two right-angled triangles. Thus the general properties of triangles involve those of all figures. THEOREM. 208. Two triangles, which have an angle of the one equal to an angle of the other, and the sides about these angles proportional, are similar. Demonstration. Let the angle AD (fig. 122), and let AB: DE: AC: DF, the triangle ABC is similar to the triangle DEF. Take AG = DE, and draw GH parallel to BC, the angle AGH =ABC (76); and the triangle AGH will be equiangular with the triangle ABC; whence AB: AG :: AC:AH; but, by hypothesis, AB: DE:: AC: DF, and, by construction, AG DE; therefore AH- DF. The two triangles AGH, DEF, have the two sides and the included angle of the one respectively equal to two sides and the included angle of the other; they are consequently equal. But the triangle AGH is similar to ABC; therefore DEF is also similar to ABC. Fig. 123 THEOREM. 209. Two triangles, which have the sides of the one parallel, or which have them perpendicular, to those of the other, each to each, are similar. Demonstration. and BC to EF, the 1. If the side AB(fig. 123) is parallel to DE, angle ABC will be equal to DEF (79): if, moreover, AC is parallel to DF, the angle ACB will be equal to DFE, and also BAC to EDF; therefore the triangles ABC, DEF, are equiangular, and consequently similar. 2. Let the side.DE (fig. 124) be perpendicular to AB, and the Fig. 124. side DF to AC. In the quadrilateral AIDH the two angles I, H, will be right angles, and the four angles will be together equal to four right angles (65); therefore the two remaining angles IAH, IDH, are together equal to two right angles. But the two angles EDF, IDH, are together equal to two right angles; consequently the angle EDF is equal to LAH or BAC. In like manner, if the third side EF is perpendicular to the third side BC, it may be shown that the angle DFE = C, and DEF=B; therefore the two triangles ABC, DEF, which have the sides of the one perpendicular to those of the other, each to each, are equiangular and similar. 210. Scholium. In the first of the above cases the homologous sides are the parallel sides, and in the second the homologous sides are those which are perpendicular to each other. Thus in the second case, DE is homologous to AB, DF to AC, and EF to BC. The case of the perpendicular sides admits of the two triangles being differently situated from those represented in figure 124; but the equality of the respective angles may always be proved, either by means of quadrilaterals, such as AIDH, which have two right angles, or by comparing two triangles, which, beside the vertical angles, have each a right angle; or we can always suppose, within the triangle ABC, a triangle DEF, the sides of which shall be parallel to those of the triangle to be compared with ABC, and then the demonstration will be the same as that given for the case of figure 124. THEOREM. 211. Lines AF, AG, &c. (fig. 125), drawn at pleasure through Fig. 125. the vertex of a triangle, divide proportionally the base BC and its parallel DE, so that DI: BF:: IK : FG :: KL : GH, &c. Demonstration. Since DI is parallel to BF, the triangles ADI, ABF, are equiangular, and DI: BF:: AI: AF; also, IK being parallel to FG, AI:AF::IK: FG; hence, on account of the common ratio, AI: AF, DI : BF :: IK: FG. It may be shown, in like manner, that IK : FG :: KL: GH, &c.; therefore the line DE is divided at the points I, K, L, as the base BC is at the points F, G, H. 212. Corollary. If BC should be divided into equal parts at the points F, G, H, the parallel DE would be divided likewise into equal parts at the points I, K, L. Fig. 126. THEOREM. 213. If, from the right angle A (fig. 126) of a right-angled triangle, the perpendicular AD be let fall upon the hypothenuse; 1. The two partial triangles ABD, ADC, will be similar to each other and to the whole triangle ABC; 2. Each side AB or AC will be a mean proportional between the hypothenuse BC and the adjacent segment BD or DC; 3. The perpendicular AD will be a mean proportional between the two segments BD, DC. Demonstration. 1. The triangles BAD, BAC, have the angle B common; moreover the right angle_BDA=BAC; consequently the third angle BAD of the one is equal to the third angle C of the other, and the two triangles are equiangular and similar. It may be demonstrated, in the same manner, that the triangle DAC is similar to the triangle BAC; therefore the three triangles are equiangular and similar. 2. Since the triangle BAD is similar to the triangle BAC, their homologous sides are proportional. Now, the side BD in the smaller triangle is homologous to the side BA in the larger, because they are opposite to the equal angles, BAD, BCA; the hypothenuse BA of the smaller is homologous to the hypothenuse BC of the larger; therefore each of the sides AB, AC, is a mean proportional be- BD: AD :: AD: DC; therefore the perpendicular AD is a mean proportional between the segments BD, DC, of the hypothenuse |