the other extremity H is given; and the straight line EAF,30 Dat. which is drawn through the given point H, parallel to BC given in position, is therefore given in position. PROPOSITION XXXVIII. If a straight line be drawn from a given point to two parallel straight lines given in position, the ratio of the segments between the given point and the parallels shall be given. Let the straight line EFG be drawn from the given point E, to the parallels AB, CD: the ratio of EF to EG is given. From the point E, draw EHK perpendicular to CD; and because from a given point E, the straight line EK is drawn to CD which is given in position, in a given angle EKC, EK is given. in position *; and AB, CD are given in position; therefore the points H, K are given: and the point E is given; wherefore EH, EK are given in magnitude, and the ratio of them is therefore given. But as EH to EK, so is EF to EG, because AB, CD are parallels; therefore the ratio of EF to EG is given. F A FH B B From the given point A, let the straight line AED be drawn to the two parallel straight lines FG, BC, and let the ratio of G A FH B B E PROPOSITIONS XXXIX. (35, 36.) If the ratio of the segments of a straight line between a given See N. point in it and two parallel straight lines be given; if one of the parallels be given in position, the other is also given in position. F E G the segments AE, AD, be given: if one of the parallels BC be given in position, the other FG is also given in position. 31 Dat. (34.) ⚫ 33 Dat. 28 Dat. 29 Dat. 1 Dat. * 33 Dat. • 28 Dat. * 29 Dat. • 2 Dat. • 31 Dat. ( 37, 38.) See N. * 33 Dat. * 28 Dat. * 29 Dat. From the point A, draw AH perpendicular to BC, and let it meet FG in K; and because AH is drawn from the given point A, to the straight line BC given in position, and makes a given angle AHD, AH is given * in position; and BC is likewise given in position, therefore the point H is given *: the point A is also given; wherefore AH is given in magnitude *, and because FG, BC are parallels, as AE to AD, so is AK to AH; and the ratio of AE to AD is given, wherefore the ratio of AK to AH is given; but AH is given in magnitude, therefore* AK is given in magnitude; and it is also given in position, and the point A is given: wherefore the point K is given. And because the straight line FG is drawn through the given point K, parallel to BC which is given in position, therefore FG is given in position. * # PROPOSITION XL. If the ratio of the segments of a straight line into which it is cut by three parallel straight lines be given; if two of the parallels be given in position, the third is also given in posi tion. Let AB, CD, HK be three parallel straight lines, of which AB, CD are given in position; and let the ratio of the segments GE, GF, into which the straight line GEF is cut by the three parallels, be given: the third parallel HK is given in position. In AB take a given point L, and draw LM perpendicular to CD, meeting HK in N: because LM is drawn from the given point L, to CD which is A E L H G NK given in position, and H G N K A E LB CF CF MD to GF is given, and as GE to GF, so is NL to NM; the ratio Cor. 6. or of NL to NM is given; and therefore the ratio of ML to 7 Dat. *Cor. 6. or 7 Dat. * 2 Dat. *30 Dat. LN is given; but ML is given in magnitude, wherefore* LN is given in magnitude: and it is also given in position, and the point L is given, wherefore the point N is given: and because the straight line HK is drawn through the given point N, parallel to CD which is given in position, therefore HK is given in position *. PROPOSITION XLI. Let the parallel straight lines AB, CD, EF, given in position, be cut by the straight line GHK: the ratio of GH to HK is given. (F.) If a straight line meet three parallel straight lines which are See N. given in position, the segments into which they cut it have a given ratio. PROPOSITION XLII. A GL B In AB take a given point L, and draw LM perpendicular to CD, meeting EF in N ; therefore LM is given in position: and CD, EF are given in position, wherefore the points M, N are given: and the point L is given; therefore the straight lines LM, MN are given in magnitude; and the ratio of LM to MN is therefore given: but as LM to MN, so is GH to HK; where- 1 Dat. fore the ratio of GH to HK is given. EK • 31 Dat. D ⚫ 33 Dat. 29 Dat. ( 39.) If each of the sides of a triangle be given in magnitude, the See N. triangle is given in species. Let each of the sides of the triangle ABC be given in magnitude: the triangle ABC is given in species. Make a triangle * DEF, the sides of which are equal, each 22. 1. to each, to the given straight lines AB, BC, CA; which can be done, because any two of them must be greater than the third; and let DE be equal to AB, B C E * EF to BC, and FD to CA; and because the two sides ED, DF are equal to the two BA, AC, each to each, and the base EF equal to the base BC; the angle EDF is equal to the angle BAC; therefore, because the angle EDF, which is equal to the angle BAC, has been found, the angle BAC is given*: in like manner the⚫1 Def. angles at B, C, are given. And because the sides AB, BC, CA are given, their ratios to one another are given *; there- *1 Dat. fore the triangle ABC is given in species. • 3 Def. 8. 1. ( 40. ) * 23. 1. * 32 Dat. • 29 Dat. * 42 Dat. * 4.6. 1 Def. 6. ( 41.) 32 Dat. PROPOSITION XLIII. If each of the angles of a triangle be given in magnitude, the triangle is given in species. Let each of the angles of the triangle ABC be given in magnitude: the triangle ABC is given in species. A Take a straight line DE given in position and magnitude, and at the points D, E, make * the angle EDF equal to the angle BAC, and the angle DEF equal to ABC; therefore the other angles EFD, BCA are equal, and each of the angles at the points A, B, C, is given, wherefore each of those at the points DEF is given. And because the straight line FD is drawn to the given point D, in DE which is given in position, making the given angle EDF, therefore DF is given in position *; in like manner EF also is given in position; wherefore the point F is given: and the points D, E are given; therefore each of the straight lines DE, EF, FD, is given in magnitude; wherefore the triangle DEF is given in species; and it is similar to the triangle ABC; which therefore is given in species. PROPOSITION XLIV. If one of the angles of a triangle be given, and if the sides about it have a given ratio to one another, the triangle is given in species. Let the triangle ABC have one of its angles BAC given, and let the sides BA, AC, about it, have a given ratio to one another: the triangle ABC is given in species. Take a straight line DE given in position and magnitude, and at the point D, in the given straight line DE, make the angle EDF equal to the given angle BAC : wherefore the angle EDF is given; and because the straight line FD is drawn to the given point D in ED, which is given in position, making the given angle EDF; therefore FD is given in position *. And because the ratio of BA to AC is given, make the ratio of ED to DF the same with it, and join EF; and because the ratio of ED to DF is given, A A B and ED is given, therefore DF is given in magnitude: and 2 Dat. it is given also in position, and the point D is given, wherefore the point F is given; and the points D, E are given, wherefore DE, EF, FD are given in magnitude; and the triangle DEF is therefore given in species; and because the triangles ABC, DEF have one angle BAC equal to one angle EDF, and the sides about these angles proportionals; the triangles are similar, but the triangle DEF is given in 6. 6. species, and therefore also the triangle ABC. * B PROPOSITION XLV. (42.) If the sides of a triangle have to one another given ratios, the See N. triangle is given in species. H Let the sides of the triangle ABC have given ratios to one another: the triangle ABC is given in species. Take a straight line D given in magnitude; and because the ratio of AB to BC is given, make the ratio of D to E the same with it; and D is given, therefore* E is given. And 2 Dat. because the ratio of BC to CA is given, to this make the ratio of E to F the same; and E is given, and therefore F. And 2 Dat. because as AB to BC, so is D to E; by composition AB and BC together are to BC, as D and E to F: but as BC to CA, so is E to F; therefore, ex æquali *, as AB and BC are to CA, so are D and E to F, and AB and BC are greater * than CA; therefore D and E are greater than F. In the same manner, any two of the three D, E, F are greater than the third. Make the triangle GHK, whose sides are equal to D, E, F, so that GH be equal to D, HK to E, and KG to F; and because D, E, F are each of them given, therefore GH, HK, KG are each of them given in magnitude; therefore the triangle GHK is given in species: but as AB to BC, so is (D to E, that is) GH to HK; and as BC to CA, so is (E to F, that is) HK to KG; therefore, ex æquali, as AB to AC, so is GH to HK. Wherefore*, the triangle ABC is 5. 6. equiangular and similar to the triangle GHK; and the triangle GHK is given in species; therefore also the triangle ABC is given in species. * 42 Dat. COR. If a triangle be required to be made, the sides of C G K 30 Dat. 29 Dat. 42 Dat. • 22. 5. • 20. 1. A. 5. 22. 1. |