adding the corresponding members and observing that BE = DE, we have --2 AB+AD+DC+BC = 4AE + 4DE. 2 But 4AE 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. their bases The triangles ADE, DEC, of which the common vertex is D, have also the same altitude, and are to each other AE, EC; that is, ADE: DEC:: AE: EC. as But it has been shown that the triangle BDE = DEC; therefore, on account of the common ratio in the two proportions (11), AD: DB:: AE : EC. 197. Corollary 1. position (IV) or also We obtain from the above theorem by com AD+DB: AD: AE+ EC: AE, ᎯᏴ : ᎯᎠ :: ᎯᏟ : ᏁᎬ, AB: BD :: AC: CE. 198. Corollary 11. If between two straight lines AB, CD, 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, OE: EG OF: FH, or OE: OF:: EG: FH; therefore, on account of the common ratio OE: OF, these two proportions give It AE: CF:: EG: FH. 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, But, by hypothesis, AD: DB:: AE: EC; consequently AO: OC: AE : EC, 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. 117. 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. 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 (67); 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, ABC DCE, 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 (67). Also, since the angle ABC= DCE, the line AB is parallel to DC; therefore the figure ACLF 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 BC: CE:: AC: DE. From these two proportions, which contain the same ratio BC: CE, we have 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 cqual, 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. Fig. 120. 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. We see, for example, that by drawing EF (fig. 121) Fig. 121. parallel to BC, the angles of the quadrilateral AEFD are equal 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 further 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 A =D (fig. 122), and let Fig. 122. AB: DE:: AC: DF, the triangle ABC is similar to the triangle DEF. Fig. 123. Take AG = DE, and draw GH parallel to BC, the angle AGH = ABC (67); 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. 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. 1. If the side AB (fig. 123) is parallel to DE, and BC to EF, the angle ABC will be equal to DEF (70); if |