sum of their squares will be greater than the square of the opposite side; if obtuse, it will be less. PROPOSITION XIV. THEOREM. In any triangle ABC, if a straight line AE be drawn from the vertex to the middle of the base, we shall have AB2+AC2=2AE2+2BE2. On BC, let fall the perpendicular AD: The triangle AEC (12. III.) gives AČ2=AE2+EC2-2EC × ED The triangle ABC (13. III.) gives AB2-AE2+EB2+2EBXED Hence, by adding the corresponding sides together, and observing that EB and EC are equal, we have ÁB2+AC2=2ÃE2+2EB2 Cor. Hence, in every parallelogram the squares of the sides are together equal to the squares of the diagonals. For the diagonals AC, BD bisect each other (32. 1.); consequently the triangle ABC gives AB2+BC2=2AE2+2BE2 The triangle ADC gives, in like manner, B E Adding the corresponding members together, and observing that BE and DE are equal, we shall have AB2+AD2+DC2+BC2—4AE2+4DE2. But 4AE2 is the square of 2AE, or of AC; 4DE is the square of BD: hence the squares of the sides are together equal to the squares of the diagonals. PROPOSITION XV. THEOREM. The line DE, drawn parallel to the base of the triangle ABC, divides the sides AB, BC proportionally; so that we have AD: DB:: AE: EC. Join BE and DC. The two triangles BDE, DEC having the same base DE, and the same altitude, since both their vertices lie in a line parallel to the base, are equivalent (2. III.). The triangles ADE, BDE, whose common vertex is E, have the same altitude, and (6. 3.) are to each other as their bases: hence we have ADE BDE :: AD: DB. : The triangles ADE, DEC, whose common B D vertex is D, have also the same altitude, and are to each other as their bases; hence ADE: DEC:: AE: EC. But the triangles BDE, DEC are equal; and therefore, since those proportions have a common ratio, we obtain AD: DB:: AE: EC. Cor. 1. Hence, componendo, we have AD + DB: AD :: AE+EC: AE, or AB: AD: : AC: AE; and also AB: BD :: AC: CE. Cor. 2. If between two straight lines AB, CD, any number of parallels AC, EF, GH, BD, &c. be drawn, those straight lines will be cut proportionally, and we shall have AE: CF :: EG: FH:: GB: HD. For, let O be the point where AB and CD meet. In the triangle OEF, the line AC being drawn parallel to the base EF, we shall have OE: AE:: OF: CF, or OE: OF:: AE: CF. In the triangle OGH, we shall likewise have OE: EG:: OF: FH, or OE OF :: EG: FH. And by reason of the common ratio OE: OF, those two proportions give AE: CF:: EG: FH. It B may be proved in the same manner, that EG : FĤ :: GB: HD, and so on; hence : G C H the lines AB, CD are cut proportionally by the parallels AC, EF, GH, &c. PROPOSITION XVI. THEOREM. Conversely, if the sides AB, AC, are cut proportionally by the line DE, so that we have AD: DB:: AE : EC, the line DE will be parallel to the base BC. D For if DE is not parallel to BC, suppose that DO is. Then, by the preceding theorem, we shall have AD: BD:: AO: OC. But by hypothesis, we have AD: DB:: AE: EC: hence we must have AO: OC:: AE: EC, or AO: AE :: OC: EC; an impossible result, since AO, the one antecedent, is less than its consequent AE, and OC, the other antecedent, is greater than its consequent EC. Hence the parallel to BC, drawn from the point D, cannot differ from DE; hence DE is that parallel. B Scholium. The same conclusion would be true, if the proportion AB AD :: AC: AE were the proposed one. For this proportion would give us AB-AD: AD:: AC—AE : AE, or BD: AD :: CE: AE. PROPOSITION XVII. THEOREM. The line AD, which bisects the angle BAC of a triangle, divides the base BC into two segments BD, DC proportional to the adjacent sides, AB, AC; so that we have BD: DC :: AB: AC. Through the point C, draw CE E parallel to AD till it meet BA produced. In the triangle BCE, the line AD is parallel to the base CE; hence (15. III.) we have the proportion, BD: DC::AB: AE. But the triangle ACE is isosceles : for since AD, CE are parallel, we B have the angle ACE DAC, and the angle AEC = BAD (23. I.); and by hypothesis DAC-BAD; hence the angle ÀCE=AEC, and consequently (13. I.) AE=AC. In place of AE in the above proportion, substitute AC, and we shall have BD : DC :: AB : AC. PROPOSITION XVIII. THEOREM. Two equiangular triangles have their homologous sides proportional, and are similar. Let ABC, CDE be two triangles which have their angles equal each to each, namely, BAC-CDE, ABC=DCE and ACB E DEC; then the homologous sides, or the sides adjacent to the equal angles, will be proportional, so that we shall have BC CE:AB: CD:: AC: DE. Place the homologous sides BC, CE in the same straight line; and produce the sides BA, ED till they meet in F. B C Since BCE is a straight line, and the angle BCA is equal to CED, it follows (23. I.) that AC is parallel to DE. In like manner, since the angle ABC is equal to DCE, the line AB is parallel to DC. Hence the figure ACDF is a parallelogram. In the triangle BFE, the line AC is parallel to the base FE; hence (15. III.) we have BC: CE:: BA: AF; or, putting CD in the place of its equal AF, BC CE: BA: CD. In the same triangle BEF, if BF be considered as the base, CD is parallel to it; and we have the proportion BC: CE :: FD: DE; or putting AC in the place of its equal FD, BC CE:: AC: DE. And finally, since both those proportions contain the same ratio BC: CE, we have AC: DE: BA: CD. Thus the equiangular triangles BAC, CDE have their homologous sides proportional. But according to the second Definition, two figures are similar when they have their angles respectively equal, and their homologous sides proportional; consequently the equiangular triangles BAC, CDE, are two similar figures. Cor. For the similarity of two triangles it is enough that they have two angles equal each to each; since the third will also be equal in both, and the two triangles will be equiangular. Scholium. Observe that, in similar triangles, the homologous sides are opposite to the equal angles; thus the angle ACB being equal to DEC, the side AB is homologous to DC; in like manner, AC and DE are homologous, because opposite to the equal angles ABC, DCE. When the homologous sides are determined, it is easy to form the proportions: AB: DC:: AC: DE:: BC: CE. PROPOSITION XIX. THEOREM. Two triangles, which have their homologous sides proportional, are equiangular and similar. Suppose we have BC: EF:: AB: DE: AC: DF; then will the triangles ABC, DEF have their angles equal, namely, A=D, B=E, C=F. At the point E, make the angle FEG B, and at F, the angle EFG C; the third G will be equal B D 14 E to the third A, and the two triangles ABC, EFG will be equiangular. Therefore, by the last theorem, we shall have BC: EF:: AB: EG; but by hypothesis, we have BC: EF :: AB: DE; hence EG-DE. By the same theorem, we shall also have BC: EF:: AC: FG; and by hypothesis, we have BC: EF :: AC: DF; hence FG=DF. Hence (11. I.) the triangles EGF, DEF, having their three sides respectively equal, are themselves equal. But by construction, the triangles EGF and ABC are equiangular: hence DEF and ABC are also equiangular and similar. E D Scholium 1. By the last two propositions, it appears that in triangles, equality among the angles is a consequence of proportionality among the sides, and conversely; so that one of those conditions sufficiently determines the similarity of two triangles. The case is different with regard to figures of more than three sides: even in quadrilaterals, the proportion between the sides may be altered without altering the angles, or the angles be altered without altering the proportion between the sides; and thus proportionality among the sides cannot be a consequence of equality among the angles of two quadrilaterals, or vice versa. It is evident, for example, that by drawing EF parallel to BC, the angles of the quadrilateral AEFD, are made equal to those of ABCD, though the proportion between the sides is different; and, in like manner, without changing the four sides AB, BC, CD, AD, we can make the point B approach D or recede from it, which will change the angles. Scholium 2. The two preceding propositions, which in strictness form but one, together with that relating to the square of the hypotenuse, are the most important and fertile in results of any in geometry: they are almost sufficient of themselves for every application to subsequent reasoning, and for solving every problem. The reason is, that all figures may be divided into triangles, and any triangle into two right-angled triangles. Thus the general properties of triangles include, by implication, those of all figures. PROPOSITION XX. THEOREM. Two triangles which have an equal angle included between proportional sides, are similar. Let the angles A and D be equal; if we have AB: DE:: AC: DF, the A triangle ABC is similar to DEF. Take AG-DE, and draw GH parallel to BC. The angle AGH (23. I.) G will be equal to the angle ABC; and the triangles AGH, ABC will be equiangular: hence we shall have AB: AG B H :: AC: AH. But by hypothesis, we have AB : DE :: AC: DF; and, by construction, AG-DE: hence AH-DF. The two triangles AGH, DEF have an equal angle included between |