therefore (V. 11.) as the triangle BAC to BAD, so is the triangle EAD to BAD; that is, the triangles BAC, EAD have the same ratio to BAD : wherefore (V. 9.) the triangles ABC, ADE are equal.* Therefore, equal triangles, &c. OTHERWISE. 1. Let CA, AD be placed in one straight line and contiguous, and let the triangles be on opposite sides of CAD, then (I. 14. cor.) BA and AE are also in one straight line. Join BD and CE; and, because the triangles ABC, ADE are equal, if BAD be added to each, the whole BCD is equal (I. ax. 2.) to the whole BED; and therefore (I. 39.) BD is parallel to CE, and (VI. 2. cor.) CA: AD:: EA: AB. 2. The same construction being made, since CA:AD::EA: AB, BD is parallel (VI. 2. cor.) to CE; and (I. 37.) the triangles BCD, BED are equal to one another : from each of them take ABD, and the remaining triangles ABC, ADE are equal. + PROP. XVI. THEOR. If four straight lines be proportionals, the rectangle contained by the extremes is equal to that contained by the means: and (2.) if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportionals. 1. Let AB : CD :: E:F; the rectangle A B.F is equal to the rectangle CD.E. Draw (I. 11.) AG, CH perpendicular to AB, CD; make AG equal to F, and CH equal to E: and complete the parallelograms BG, DH. Because AB: CD::E:F; and that E is equal to CH, and F to AG; therefore (V. 7.) AB:CD::CH: AG. Hence the sides of the parallelograms BG, DH, about the equal angles are reciprocally F H 그 • It is evident from the second corollary to the thirty-eighth proposition of the first book, that this proposition is true as well when the angles are supplemental, as when they are equal. † From either this very easy method of demonstrating the 15th proposition, or from the method given in the text, the proof of the 14th would follow immediately, by means of the 34th of the first book, if the diagonals of the parallelograms were drawn. It may be remarked that a proof similar to the second one given above, would be obtained by placing the triangles so as to have their equal angles coinciding, instead of being vertically opposite. In the figure, the straight line CE, which is not actually drawn, may be conceived to be drawn. E н, A B proportional : wherefore (VI. 14.) the parallelogram BG is equal to DH. Now the parallelogram BG is contained by AB and F, because AG is equal to F; and DH is contained by CD and E, because CH is equal to E. Therefore the rectangle AB.F is equal to the rectangle CD.E. 2. If the rectangle AB.F be equal to the rectangle CD.E; then AB : CD :: E: F. The same construction being made, because the rectangle AB.F is equal to the rectangle CD.E, and that the rectangle BG is contained by AB and F, because AG is equal to F; and the rectangle DH by CD and E, because CH is equal to E; therefore the parallelograms BG, DH are equal; and they are equiangular. But (VI. 14.) the sides about the equal angles of equal parallelograms are reciprocally proportional : wherefore, as AB : CD::CH: AG; or (const.) AB : CD :: E: F. * Therefore, if four straight lines, &c. PROP. XVII. THEOR. If three straight lines be proportionals, the rectangle contained by the extremes is equal to the square of the mean : and (2.) if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals. Let the three straight lines A, B, C be proportionals, viz., A:B :: B: C; the rectangle A.C is equal to the squate of B. 1. Take D equal to B; and because A :B::B:C, and that B is equal to D; therefore A:B::D:C. But (VI. 16.) if four straight lines be proportionals, the rectangle under the extremes is equal to that under the means: there. A. fore the rectangle A.C is equal to the rectangle B.D. But the rectangle B.D is the square of B; because DB is equal to D: therefore the rectangle A.C is equal to the square of B. 2. If the rectangle A.C be equal to the square of B; A:B :: B: C. The same construction being made, because the rectangle A.C is equal to the square of B, and the square of B is equal to the B This proposition, of which the one following it is a case, affords the means of deriving the equality of rectangles, and the proportion of straight lines, from one another. It evidently corresponds to V. Sup. prop. 1. cor. and prop. 2. cor. 1., and it might be regarded as an immediate result of those corollaries, without any distinct proof, if the lines were expressed (1. 46. cor. 4.) by lineal units, and the rectangles by superficial ones. This proposition and the fourth of this book, when employed in connexion with one another, form one of the most powerful instruments in geometrical investigations : and they facilitate in a peculiar degree the application of algebra to such inquiries. rectangle B.D, because B is equal to D; therefore the rectangle A.C is equal to the rectangle B.D. But (VI. 16.) if the rectangle of the extremes be equal to that of the means, the straight lines are proportionals: therefore A:B::D:C; but B is equal to D; wherefore as A :B::B:C. Therefore, if three straight lines, &c. PROP. XVIII. PROB. H F E Upon a given straight line to describe a figure similar to a given rectilineal figure, and such that the given line shall be homologous to an assigned side of the given figure. * Let AB be the given straight line, on which it is required to describe a rectilineal figure similar to a given rectilineal figure, and such that AB may be homologous to CD, a side of the given figure. First, let the given rectilineal figure be the triangle CDE. Make the angles BAF, ABF respectively equal to DCE, CDE; and (V1. 4. cor.) the triangle ABF is similar to CDE, and has AB homologous to CD. Again, let the given figure be the quadrilateral CDGE. DE, and, as in the first part, describe the triangle ABF having the angles BAF, ABF respectively equal to DCE, CDE, and also the tri. angle BFH having the angles FBH, BFH respectively equal to EDG, DEG. Then (1.32. cor. 5.) the angles BHF, DGE are equal, and const.) A and C are equal. Also since (const.) ABF, FBH are respectively equal to CDE, EDG, the whole ABH is equal to the whole CDG. For the same reason, AFH is equal to CEG; and there. fore the quadrilateral figures ABHF, CDGE are equiangular. But likewise these figures bave their sides about the equal angles proportional. For the triangles ABF, CDE being equiangular. and also BFH, DEG; as BA : AF:: DC : CE; and as FH: HB :: EG: GD. Also, in the same triangles as AF: FB :: CE: ED; and as FB : FH :: ED: EG; therefore, ex æquo, AF: FH:: CE: EG. In the same manner it may be that AB : BH :: CD: DG: wherefore (VI. def. 1.) the figures ABHF, CDGE are similar to one another. Next, let the given figure be CDKGE. Join DG; and, as in the second case, describe the figure ABHF, similar to CDGE, and similarly situated : also, describe the triangle BHL baving K A D proved * The latter condition may be also expressed by saying, that the two figures are to be “ similarly situated” on the two straight lines AB and CD. H G E K А D the angles BHL, HBL respectively equal to DGK, GDK. Then (1. ax. 2.) the whole angles FHL, ABL are respectively equal to the whole angles EGK, CDK; and (I. 32. cor. 5.) the angles L and K are equal. Therefore the figures ABLHF, CĎKGE are equiangular. Again, because the quadrilaterals ABHF, CDGE, and the triangles BLH, DKG are similar, as FH: HB :: EG : GD, and HB: HL :: GD: GK; therefore, ex æquo, FH:HL:: EG: GK. In like manner, it may be shown, that AB : BL :: CD: DK: and because the quadrilaterals ABHF, CDGE, and the triangles BLH, DKG are similar, the sides about the angles A and C, L and K, AFH and CEG are proportional. Therefore (VI. def. 1.) the five-sided figures ABLHF, CDKGE are similar, and the sides AB and CD are homologous. In the same manner a rectilineal figure of six or more sides may be described on a given straight line, similar to one given : which was to be done. * PROP. XIX. THEOR. Similar triangles are to one another in the duplicate ratio of their homologous sides. Let ABC, DEF be similar triangles, having the angles B and E equal, and AB: BC::DE: EF, so that (V. def. 13.) the side BC is homologous to EF. The triangle ABC has to the triangle DEF, the duplicate ratio of that which BC has to EF. Take (VI. 11.) BG a third proportional to BC, EF, so that AB: BC :: DE: EF; alternately, as BC: EF::EF:BG; therefore (V. 11.) as AB:DE:: EF:BG. The sides, therefore, of the triangles ABG, DEF, which are about the equal angles, are reciprocally proportional ; and therefore (VI. 15.) the triangles ABG, DEF are equal. Again, because BC: EF:: EF: BG; and that if three straight lines be proportionals, the first is said D B G H C • In practice, if A B be parallel to CD, the construction is most easily effected by drawing A F and BF parallel to CE and DE; then FH and BH parallel to EG and DG; and, lastly, HL and BL parallel to GK and DK. The doing of this is much facilitated by employing the useful instrument, the parallel ruler. For the easiest methods, however, of performing this and many other problems, the student inust have recourse to works that treat expressly on such subjects, particularly treatises on practical geometry, surveying, and the use of mathematical instruments. + The third proportional might be taken to EF and BC, and placed from E along E F produced ; and then a triangle equal to ABC would be formed by joining D with the extremity of the produced line. (V. def. 11.) to have to the third the duplicate ratio of that which it has to the second; BC.therefore has to BG the duplicate ratio of that which BC has to EF. But (VI. 1.) as BC to BG, so is the triangle ABC to ABG. Therefore (V. 11.) the triangle ABC has to ABG the duplicate ratio of that which BC has to EF. But the triangle ABG is equal to DEF; wherefore also the triangle ABC has to DEF the duplicate ratio of that which BC has to EF. * Therefore, similar triangles, &c. Cor. From this it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any triangle upon the first to a similar triangle, similarly described on the second. PROP. XX. THEOR. SIMILAR polygons may be divided into the same number of similar triangles : (2.) these triangles have the same ratio to one another that the polygons have : and (3.) the polygons have to one another the duplicate ratio of that which their homologous sides have. Let AD, FK be similar polygons, and let the sides AB, FG be homologous: the polygons may be divided into the same number of similar triangles : (2.) these triangles have, each to each, the same ratio which the polygons have : and (3.) the polygon AD has to the polygon FK the duplicate ratio of that which the side AB has to FG. 1. Let the angles AED, FLK be equal, and the sides about them proportional; and draw the diagonals EB, EC, LG, LH. Then, (VI. def. 1.) because the polygons are similar, the angles A and F are equal, and BA : AE::GF:FL: where. fore, because the triangles ABE, FGL have an angle in one equal to an angle in the other, and the sides about those angles proportionals, the triangle ABE is (VI. 6.) equiangular, and therefore (VI. 4. and def. 1.) similar to the triangle FGL; wherefore the angles A BE, FGL are equal. Again, because the polygons are similar, the whole angle ABC is equal (VI. def. 1.) to the whole FGH; therefore (I. ax. 3.) the remaining angles EBC, LGH are equal : + and (VI. 4.) because the This proposition might also be proved by making BH equal to EF, joining AH, and drawing through H a parallel to AC. The triangle cut off by the parallel is equal (1. 26.) to DEF. But (VI. 1.) the triangle ABC is to the triangle ABH, as BC to BH; and, for the same reason, the triangle ABH is to the triangle cut off by the parallel, or to DEF, as BC to BH, or (const.) as BH to BG. Therefore, ex æquo, the triangle ABC has to the triangle DEF, the same ratio that BC has to BG, or (V. def. 11.) the duplicate ratio of that which BC bas to EF. + Should the angles A and F be re-entrant, so that the points A and F would be on the other sides of A B and LG, the angles EBC, LGH would not be the difference, but the sum of ABC, ABE, and of FGH, FGL. M F B E G L D H |