greater than the Angle B A C ; but the Angle BDC has been proved to be greater than the Angle CED. Wherefore the Angle BDC fhall be much greater than the Angle BAC. And fo, if two Right Lines be drawn from the extreme Points of one Side of a Triangle to any Point within the fame, thefe two Lines fhall be less than the other two Sides of the Triangle, but contain a greater Angle; which was to be demonftrated. PROPOSITION PROBLEM. XXII. To defcribe a Triangle of three Right Lines, which are equal to three others given: But it is requifite, that any two of the Right Lines taken together be greater than the third; because two Sides of a Triangle, bowfoever taken, are together greater than the third Side. IET A, B, C, be three Right Lines given, two of which, any ways taken, are greater than the third; viz. A and B together greater than C; A and C greater than B; and B and C greater than A. Now it is required to make a Triangle of three Right Lines equal to A, B, C: Let there be one Right Line D E, terminated at D, but infinite towards E; and take* DF * 3 of this. equal to A, FG equal to B, and G H equal to C; and about the Centre F, with the Distance FD, defcribe a Circle DK Lt; and about the Centre G, † 3 Poß. with the Distance G H, defcribe another Circle KLH, and join KF, KG. I fay, the Triangle KFG is made of three Right Lines, equal to A, B, C; for, because the Point F is the Centre of the Circle D KL, FK fhall be equal to FDt: But FD is equal to A; therefore F K is also equal to A. Again, because the Point G is the Centre of the Circle L KH, GK will be equal to GH: But G H is equal to C; Def. 15therefore fhall G K be alfo equal to C: But F G is likewife equal to B; and confequently the three Right Lines KF, FG, KG, are equal to the three Right Lines A, B, C. Wherefore, the Triangle K FG is made of three Right Lines KF, F G, KĜ, equal to the three given Lines A, B, C, which was to be done. C 3 PRO PROPOSITION XXIII. PROBLEM. With a given Right Line, and at a given Poins in it, to make a Right-lin'd Angle equal to a Right-lin❜d Angle given. ET the given Right Line be A B, and the Point given therein A, and the given Right-lin❜d Angle DCE. It is required to make a Right-lin❜d Angle at the given Point A, with the given Right Line AB, equal to the given Right-lin'd Angle DCE. Affume the Points D and E at Pleasure in the Lines CD, CE, and draw DE; then, of three Right * 22 42 of this. Lines equal to CD, DE, EC, make a Triangle AFG, fo that A F be equal to CD, AG to CE, and FG to DE. Then, because the two Sides D'C, CE, are equal to the two Sides F A, A G, each to each, and the Bafe DE equal to the Bafe F G; the Angle DCE fhall + 8 of this. be + equal to the Angle FA G. Therefore, the Rightlin'd Angle FAG is made, at the given Point A, in the given Right Line AB, equal to the given Right-lin'd Angle DCE; which was to be done. PROPOSITION XXIV. THEOREM. If two Triangles bave two Sides of the one equal to two Sides of the other, each to each, and the Angle of the one contained under the equal Right Lines, greater than the correfpondent Angle of the other; then the Bafe of the one will be greater than the Bafe of the other. ET there be two Triangles A B C, DEF, having two Sides A B, A C, equal to the two Sides DE, DF, each to each, viz. the Side A B equal to the Side D E, and the Side AC equal to D F; and let the Angle B A C be greater than the Angle EDF. fay, the Bafe BC is greater than the Bafe E E. For 23 of thin. For because the Angle BAC is greater than the Angle EDF; make an Angle E DG at the Point Din the Right Line D E,equal to the Angle B AC; and make + D G equal to either AC or D F, and + 3 of this. join EG, FG. Cafe 1. When EG falls above E F; then, becaufe AB is equal to D E, and AC to DG, the two Sides BA, AC, are each equal to the two Sides E D, DG, and the Angle BAC equal to the Angle EDG: Therefore the Bafe BC is equal to the Bafe E G. 14 of this. Again, because DG is equal to D F, the Angle D F G is equal to the Angle DGF; and fo the Angle† 5 of tbis. DFG is greater than the Angle E GF: And confequently the Angle EFG is much greater than the Angle EGF. And because EFG is a Triangle, having the Angle E F G greater than the Angle EGF; and the greatest Angle fubtends the greatest* 19 of thin. Side, the Side EG fhall be greater than the Side EF. But the Side EG is equal to the Side BC: Whence BC is likewife greater than E F. Cafe 2. When EG falls upon EF; then E G is greater than E F: And confequently BC is greater than EF. Ax. 5• Cafe 3. When EG falls below EF; then DG, EG, are together greater than DF and EF toge-1 21 of this ther, and by taking away the Equals DG, DF, there remain EG greater than EF. Therefore BC, which is equal to E G, will be alfo greater than EF. Therefore, if two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Angle of the one contained under the equal Right Lines, greater than the correspondent Angle of the other; then the Bafe of the one will be greater than the Bafe of the other which was to be demonftrated. PROPOSITION XXV. THEOREM. If two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Bale of the one greater than the Bafe of the other; they shall also have the Angles contained by the equal Sides, the one greater than the other. LEF there be two Triangles ABC, DEF, having two Sides A B, A C, each equal to two Sides DE, DF, viz. the Side A B equal to the Side DE, and the Side AC to the Side DF; but the Bafe BC greater than the Bafe EF; I fay, the Angle BAC is allo greater than the Angle EDF. * For if it be not greater, it will be either equal or lefs, But the Angle B A C is not equal to the Angle 4 of this. EDF; for if it was, the Bafe BC would be equal to the Bafe EF; but it is not: Therefore the Angle BAC is not equal to the Angle EDF, neither will 24 of bis, it be lefler; for if it should, the Bafe B C would be f less than the Base EF; but it is not. Therefore the Angle B A C is not lefs than the Angle EDF; but it has likewife been proved not to be equal to it. Wherefore the Angle BAC is neceffarily greater than the Angle EDF. If, therefore, twa Triangles have twa Sides of the one equal to two Sides of the other, each to each, and the Bafe of the one greater than the Bafe of the ather; they shall also have the Angles contained by the equal Sides, the one greater than the other; which was to be demonftrated. PRO PROPOSITION XXVI. THEOREM. If two Triangles bave two Angles of the one equal to two Angles of the other, each to each, and one Side of the one equal to one Side of the other, either the Side lying between the equal Angles, or which fubtends one of the equal Angles; the remaining Sides of the one Triangle hall be alfo equal to the remaining Sides of the other, each to his correfpondent Side; and the remaining Angle of the one equal to the remaining Angle of the other. L ET there be two Triangles A B C, DEF, having two Angles ABC, BC A, of the one, equal to two Angles DEF, EFD, of the other, each to each, that is, the Angle ABC equal to the Angle DEF, and the Angle B C A equal to the Angle EFD. And let one Side of the one be equal to one Side of the other, which first let be the Side lying between the equal Angles, viz. the Side B C equal to the Side E F. I fay, the remaining Sides of the one Triangle will be equal to the remaining Sides of the other, each to each, that is, the Side A B equal to the Side DE, and the Side AC equal to the Side DF, and the remaining Angle BAC equal to the remaining Angle EDF. For if the Side A B be not equal to the Side D E, one of them will be the greater, which let be A B make G B equal to D E, and join G C. • * Then, because B G is equal to D E, and BC to EF, the two Sides GB, BC, are equal to the two Sides DE, EF, each to each; and the Angle GBC equal to the Angle DEF. The Bafe GC is equal to the 4 of this. Bafe D F, and the Triangle GBC to the Triangle DEF, and the remaining Angles equal to the remaining Angles, each to each, which fubtend the equal Sides. Therefore the Angle GCB is equal to the Angle DF E. But the Angle D F E, by the Hypo 4 thefis, |