Scholium. The converse of this proposition, viz. triangles which are to each other as their altitudes have equal bases, is evidently true (see preceding scholium). PROPOSITION III. THEOREM. If four straight lines are proportional, the rectangle contained by the extremes is equivalent to the rectangle contained by the means, and conversely, if two rectangles are equivalent, their containing sides are proportional. Let the four lines AB, CD, DE, BF be proportional, then AF, the rectangle of the extremes, is equivalent to CE, the rectangle of the means. Make DG equal to BF, and draw GH parallel to DC. Then (Prop. I. Cor.) AF CG:: AB : CD ; but by hypothesis, AB: CD:: DE: BF: therefore (Prop. II. B. V.) AF: CG:: DE: BF, or DG, but DE: DG:: CE CG; therefore (Prop. II. B. V.) AF : CG:: CE: CG, and the consequents being equal the antecedents are equal (Prop. IX. Cor. 2. B. V.), that is, AF÷CG. Conversely. Let the rectangle AF be equivalent to the rectangle CE. Then (Prop. I. Cor.) AB: CD:: AF or CE: CG, but CE : CG :: DE: DG (Prop. I. Cor.); therefore (Prop. II. B. V.) AB: CD:: DE: DG or BF. Cor. It follows that if three lines are in continued proportion, the rectangle of the extremes is equivalent to the square of the mean, and conversely, if a square be equivalent to a rectangle, the side of the square is a mean between the sides of the rectangle. The rectangles contained by the corresponding lines which form two proportions are themselves proportional. Let there be the proportions AB: BC:: CD: DE, and BF: CG:: DH: EI, then also AF: BG :: CH : DI. G H A B B D D In CG or its production take Cf=BF, and in EI or its production take Eh-DH, and through fand h draw parallels to BC and DE. Then (Prop. I. Cor. B. VI.) AF: Bf:: AB: BC:: CD: DE: CH: Dh, and alternately, AF: CH:: Bf: Dh. Now Bf BG:: Cf: CG:: Eh: EI :: Dh : DI; hence, alternately Bf: Dh:: BG: DI; but it has just been shown that Bf: Dh :: AF: CH; therefore AF: CH;: BG: DI, or AF: BG:: CH: DI. Cor. 1. Hence the squares of four proportional lines are proportional. Cor. 2. If three lines are in continued proportion, the square on the first is to the square on the second, as the first line is to the third. Thus if A, B, C are three lines in continued proportion, then A: B:: B: C, and since A: B :: A: B, we have by the proposition, the square on A to the square on B, as the rectangle of A, B to the rectangle of B, C, and these rectangles are as A to C (Prop. I. Cor.). Scholium. The converse of this proposition is not true, for it cannot be inferred that proportional rectangles have proportional sides, since a rectangle may be transformed into an equivalent one, having a side of any given length (Prop. XVII. B. IV.). The converse of the corollaries are true, that is, first, if four squares be proportional, their sides will be proportional; for let A, B, C, D represent the sides of four proportional squares, then if these sides are not proportional let there be the proportion A: B:: C: Q, then, by the corollary, A2 : B2 :: CQ; but by hypothesis, A2: B:: C2: D2; consequently (Prop. IX. Cor. 3. B. V.), Q-D2, and therefore Q = D. Again, if the squares on two lines are to each other as the first line is to a third, the three lines are in continued proportion, for let A2 : B2 :: A : C, and let there be the continued proportion A B :: B: Q, then, by the corollary, Ao : B2 :: A: Q; but by hypothesis, A2: B:: A: C, hence (Prop. IX. Cor. 3. B. V.), Q=C. : PROPOSITION V. THEOREM. If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those produced, proportionally; and conversely, if the sides or the sides produced, be cut proportionally, the cutting line will be parallel to the third side of the triangle. In the triangle ABC let DE be drawn parallel to BC, then AD: DB:: AE: EC. For join BE, CD. Then the triangles BDE, DEC are equivalent, for their bases are the same, and being between parallels their altitudes are equal; therefore ADE: BDE:: ADE : DEC, but (Prop. I.), ADE: BDE::AD: DB, and ADE: DEC: AE: EC; hence (Prop. II. B. V.) AD: DB :: AE: EC. Conversely. Let now DE cut the sides AB, AC or their production, so that AD: DB:: AE: EC, then DE will be parallel to BC. For the same construction remaining, AD: DB :: AED : DEB, and AE: EC:: AED: DEC, hence AED: DEB:: AED DEC; consequently (Prop. IX. Cor. 2. B. V.), the triangles DEB, DEC are equivalent, and having the same base DE, their altitudes are equal, that is, they are between the same parallels. Cor. AD+DB: AD :: AE+EC: AE, that is, AB : AD :: AC: AE, also AB: BD:: AC: CE. PROPOSITION VI. THEOREM. If through two sides of a triangle two lines be drawn parallel to the third side, the portions intercepted will be to each other as the sides themselves. Let the two sides AB, AC of the triangle ABC be divided by the lines DF, EG parallel to BC; then AB AC :: DE: FG. : For by last proposition AD: AF:: DE FG, and by the corollary AD : AF:: AB: AC; consequently, AB : AC: DE: FG. D E B A Cor. Hence if any number of parallels be drawn, the sides will be cut proportionally, the opposite intercepted portions being to each other as the sides themselves. Scholium. The converse of this proposition does not follow, that is, it is not true, that if the portions intercepted by two lines cutting two sides of a triangle are to each other as those sides, the cutting lines will be parallel to the third side of the triangle; for any two lines drawn from D, E to intercept on the other side a portion equal to FG, will be drawn as this converse directs, although only a single pair can be parallel to the side BC. PROPOSITION VII. THEOREM. The line which bisects any angle of a triangle divides the opposite side into portions, which are to each other as the adjacent sides. Let AD bisect the angle A of the triangle ABC, then BD: DC:: AB: AC. Draw CE parallel to DA, meeting BA produced in E. Then (Prop. V.) BD : DC : : BA : AE. Now because AD, EC are parallel, the angle E is equal to the angle BAD, and the angle ACE to the angle CAD, which is equal, by hypothesis, to BAD; it appears then that the angles E and ACE are each equal to BAD; therefore AE is equal to AC; hence, putting AC for AE in the above proportion, we have BD: DC:: BA: AC. B E PROPOSITION VIII. THEOREM. (Converse of Prop. VII.) If a line from the vertex of any angle of a triangle divide the side opposite into portions, which are to each other as the sides adjacent, the line so drawn bisects the angle. In the triangle ABC (preceding diagram) let the line AD divide BC, so that BD: DC:: BA AC, then is the angle BAD equal to the angle CAD. Draw CE parallel to DA, meeting BA, produced in E. Then (Prop. V.) BD : DC :: BA: AE; but by hypothesis, BD: DC :: BA : AC, therefore BA : AE :: BÁ : AC; consequently, AE= AC, and therefore the angle ACE to the angle AEC; but because of the parallels AD, EC, the angles AČE, AEC are respectively equal to the angles DAC, DAB; these angles are therefore equal, and consequently AD bisects the angle BAC. If two triangles have the angles of the one respectively equal to those of the other, the sides containing the equal angles are proportional. Let the triangles ABC, DCE have the angles A, B, in the one, respectively equal to the angles D, C, in the other; these triangles will be similar. For let them be placed so that two homologous sides BC, CE may form one straight line, and produce BA, ED till they meet in F. A F E Then since the angle B is equal to the angle DCE, the line BF is parallel to the line CD; also since the angle ACB is equal to the angle E, the line AC is parallel to the line FE; therefore CF is a rhomboid, and con- B sequently AF-CD, and FDAC. Now because AC is parallel to FE, we have (Prop. V.) BC: CE: BA: AF, and because CD is parallel to BF we have BC: CE :: FD: DE, and if in these proportions CD be put for its equal AF, and AC for its equal FD, they become whence BC CE: BA: CD |