base bc, the same multiple is the triangle a hc of the triangle a bc. For the same reason, whatever multiple the base lc is of the base cd, the same multiple is the triangle a lc of the triangle a dc: and if the base hc be equal to the base cl, the triangle a hc is also equal to the triangle alc (i. 38): and if the base hc be greater than the base cl, likewise the triangle a hc is greater than the triangle alc; and if less, less : therefore, since there are four magnitudes, viz. the two bases b c, cd, and the two triangles a bc, acd; and of the base b c, and the triangle a b c, the first and third, any equimultiples whatever have been taken, viz. the base ho e h 8 d 1 and triangle a hc; and of the base cd and triangle ac d, the second and fourth, have been taken any equimultiples whatever, viz. the base cl and triangle alc; and that it has been shewn, that if the base hc be greater than the base cl, the triangle a h cis greater than the triangle alc; and if equal, equal ; and if less, less : therefore (v. def. 5), as the base b c is to the base cd, so is the triangle a b c to the triangle a cd. And because the parallelogram ce is double of the triangle a b c (i. 41), and the parallelogram cf double of the triangle a cd, and that magnitudes have the same ratio which their equimultiples have (v. 15); as the triangle a b c is to the triangle a cd, so is the parallelogram e c to the parallelogram cf; and because it has been shewn, that as the base bc is to the base cd, so is the triangle a bc to the triangle acd; and as the triangle a b c is to the triangle a cd, so is the parallelogram ec to the parallelogram cf; therefore, as the base bc is to the base cd, so is (v. 11) the parallelogram ec to the parallelogram cf. Wherefore, triangles, &c.' Q. Ê. D. Cor. From this it is plain, that triangles and parallelograms that have equal altitudes are one to another as their bases. Let the figures be placed so as to have their bases in the same straight line ; and having drawn perpendiculars from the vertices of the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are (i. 33), because the perpendiculars are both equal and parallel to one another. Then, if the same construction be made as in the proposition, the demonstration will be the same. PROPOSITION II.—THEOREM. If a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or these produced, proportionally ; and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle. LET de be drawn parallel to bc, one of the sides of the triangle a bc:bd is to da, as ce to ea. Join be, cd; then the triangle bde is equal to the triangle cde (i. 37), because they are on the same base de, and between the same parallels de, bc. But a d e is another triangle, and equal magnitudes have to the same the same ratio (v. 7) ; therefore, as the triangle bd e to the triangle a de, so is the triangle cde to the triangle a de, but as the triangle bde to the triangle ade, so is (vi. 1) bd to da, because having the same altitude, viz. the perpendicular drawn from the point e to a b, d b e b they are to one another as their bases ; and for the same reason, as the triangle cde to the triangle a de, so is ce to ea. Therefore, as b d to da, so is ce to e a (v. 11). Next, let the sides a b, a c, of the triangle a b c, or these produced, be cut proportionally in the points d, e, that is, so that bd be to da asi ce to ea, and join de; de is parallel to b c. The same construction being made, because as b d to da, so is ce to ea; and as bd to d a, so is the triangle bde to the triangle a de (vi. 1); and as ce to ea, so is the triangle cde to the triangle ade; therefore the triangle bde is to the triangle a d e, as the triangle cde to the triangle a de; that is, the triangles b de, cd e have the same ratio to the triangle ade; and therefore (v. 9) the triangle bde is equal to the triangle cde; and they are on the same base de: but equal triangles on the same base are between the same parallels (i. 39) ; therefore d e is parallel to bc. Wherefore, if a straight line, &c. Q. É. D. PROPOSITION III.-THEOREM. If the angle of a triangle be divided into two equal angles by a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another; and if the segments of the base have the same ratio which the other sides of the triangle e have to one another, the straight line drawn from the vertex to the point of section divides the vertical angle into two equal angles. LET the angle ba c of any triangle a b c be divided into two equal angles by the straight line ad : bd is to dc as b a to a c. Through the point c draw ce parallel (i. 31) to da, and let ba produced meet ce in e; because the straight line ac meets the parallels ad, ec, the angle a ce is equal to the alternate angle cad (i. 29): but cad, by the hypothesis, is equal to the angle bad; wherefore bad is equal to the angle a ce.. Again, because the straight line ba e meets the parallels a d, ec, the a outward angle bad is equal to the inward and opposite angle a ec: but the angle a ce has been proved equal to the angle bad; therefore also a ce is equal to the angle a ec, and consequently the b d side a e is equal to the side ac (i. 6): and because a d is drawn parallel to one of the sides of the triangle bce, viz. to ec, bd is to dc, as ba to a e (vi. 2), but a e is equal to ac; therefore as bd to dc, so is ba to a c (v.7). Let now bd be to dc, as b a to a c, and join a d; the angle bac is divided into two equal angles by the straight line a d. The same construction being made; because, as bd to d c, so is b a to ac; and as bd to cd so is ba to a e (vi. 2), because a d is parallel to ec; therefore ba is to a c, as ba to ae (v. 11); consequently a c is equal to a e (v. 9), and the angle a ec is therefore equal to the angle a ce (i. 5): but the angle a ec is equal to the outward and opposite angle bad; and the angle ace is equal to the alternate angle cad (i. 29) : wherefore also the angle bad is equal to the angle cad; therefore the angle bac is cut into two equal angles by the straight line ad. Therefore, if the angle, &c. Q. E. D. PROPOSITION A.—THEOREM, If the outward angle of a triangle made by producing one of its sides be divided into two equal angles by a straight line which also cuts the base produced, the segments between the dividing line and the extremities of the base have the same ratio which the other sides of the triangle have to one another; and if the segments of the base produced have the same ratio which the other sides of the triangle have, the straight line drawn from the vertex to the point of section divides the outward angle of the triangle into two equal angles. LET the outward angle ca e of any triangle a bc be divided into two qual angles by the straight line ad which meets the base produced in : bd is to dc, as b a to a c. Through c draw cf parallel to ad (i. 31); and because the straight line ac meets the parallels a d, cf, the angle a cf is equal to the alternate a angle cad (i. 29): but cad is equal to the angle dae (hyp.); therefore also dae is equal to the angle a cf. Again, because the straight line fa e meets the parallels a d, fc, the outward e angle da e is equal to the inward and opposite angle cfa: but the angle a cf has been proved equal to the angle da e; therefore also the angle a cf is equal to the angle cfa, and conse quently the side af is equal to the side b d ac (i. 6): and because a d is parallel to fc, a side of the triangle bcf, bd is to dc, as ba to af (vi. 2); but a f is equal to ac; as therefore bd is to dc, so is ba to ac. Let now bd be to do as b a to ac, and join ad; the angle cad is equal to the angle da e. The same construction being made, because bd is to dc as b a to ac; and that bd is also to dc, as ba to af (vi. 2); therefore ba is to ac as ba to af (v. 11); wherefore a c is equal to af (v. 9), and the angle afc equal (i. 5) to the angle a cf: but the angle afc is equal to the outward angle e a d, and the angle a cfto the alternate angle cad; therefore also ead is equal to the angle cad. Wherefore, if the outward, &c. Q. E. D. PROPOSITION IV.-THEOREM. The sides about the equal angles of equiangular triangles are proportionals; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios. LET a bc, dce be equiangular triangles, having the angle a b c equal to the angle dce, and the angle ach to the angle dec, and consequently (i. 32) the angle bac equal to the angle cde. The sides about the equal angles of the triangles a b c, dce are proportionals; and those are the homologous sides which are opposite to the equal angles. Let the triangle dce be placed so that its side ce may be contiguous to bc, and in the same straight line with it: and because the angles a b c, a cb, are together less than two right angles (i. 17), a b c and d ec, which is equal to a cb, are also less than two right angles; wherefore ba, ed produced shall meet (i. ax. 12); let them be produced and meet in the point f; a and because the angle a b c is equal to the angle dce, bf is parallel to cd (i. 28). Again, because the angle a cb is equal to the angle dec, ac is parallel to fe (i. 28): therefore b C e facd is a parallelogram, and consequently af is equal to cd, and ac to fd (i. 34): and because a c is parallel to fe, one of the sides of the triangle fb e, ba is to af as b c to ce (vi. 2): but a f is equal to cd; therefore (v. 7), as ba to cd, so is bc to ce; and alternately, as a b to bc, so is dc to ce. d Again, because cd is parallel to bf, as b c to ce, so is fd to de (vi. 2); but fd is equal to ac; therefore, as b c to ce, so is a c to de: and alternately, as b c to ca, so is ce to ed : therefore, because it has been proved that a b is to b c, as dc to ce, and as b c to ca, so is ce to ed, ex æquali (v. 22), ba is to a c, as cd to de. Therefore the sides, &c. Q. E. D. PROPOSITION V.-THEOREM. If the sides of two triangles about each of their angles be proportionals, the triangles shall be equiangular, and have their equal angles opposite to the homologous sides. LET the triangles a b c, def have their sides proportionals, so that a b is to bc as de to ef; and bc to ca as ef to fd; and consequently, ex æquali, ba to a c as ed to df; the triangle a bc is equiangular to the triangle d e f, and their equal angles are opposite to the homologous sides, viz. the angle a bc equal to the angle de f, and bca to efd, and also bac to ed f. At the points e, f, in the straight line ef, make (i. 23) the angle feg a 8 is equal (i. 8) to the angle gef, and the other angles to the other angles which are subtended by the equal sides (i. 4). Wherefore the angle d fe is equal to the angle gfe, and edf toegf: and because the angle def is equal to the angle gef, and gef to the angle a bc; therefore the angle a b c is equal to the angle def. For the same reason, the angle a c b is equal to the angle dfe, and the angle at a to the angle at d. Therefore the triangle a b c is equiangular to the triangle def. Wherefore, if the sides, &c. Q. E. D. to ef b PROPOSITION VI.-THEOREM. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be |