† 190f ibis. greater Angle subtends + the greater Side ; the Side D B will be greater than the Side B C. But D B is equal to B A and AC together. Wherefore the Sides BA, A C, together, are greater than the Side B C. In the same manner we demonstrate, that the sides AB, B C, together, are greater than the Side CAS and the sides BC, CA, together, are greater than the Side A B. Therefore, two sides of any Triangle, bowsoever taken, are together greater than the third Side ; which was to be demonstrated. PROPOSITION XXI. THEOREM. Points of one side of a Triangle to any Point contain a greater Angle. FOR let two Right Lines BD, DC, be drawn from the Extremes B, C, of the Side B C of the Triangle A BC, to the Point D within the same. I say, B D, DC, are less than B A, A C, the other ewo Sides of the Triangle, but contain an Angle BD C greater than the Angle B AC. For produce B D to E. Then because two Sides of every Triangle together * 20 of this. are * greater than the third, BA, A E the iwo Sides of the Triangle A B E, are greater than the Side B E. Now, add E C, which is common, and the Sides Axo 4. BA, A C, will be t greater than B E, EC. Again, because CE, E D, the two Sides of the Triangle C E D, are greater than the Side CD, add DB, which is common, and the sides CE, E B, will be greater than CD, DB. But it has been proved, that BA, A C, are greater than B E, E C. Wherefore BA, A C, are much greater than BD, DC. Again, because the outward Angle of every Triangle I is 1 16 of Ibis. greater than the inward and opposite one ; BDC, the outward Angle of the Triangle CD E, shall be greater than the Angle CED. For the same Reason, CED, the cytward Angle of the Triangle A B E, is likewise greater greater than the Angle B AC; but the Angle BDC PROBLEM. are equal to three others given : But it is requi. ET 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 Е, terminated at D, but infinite towards E; and take *DF. 3 of Ibis, cqual to A, F G, equal to B, and G H equal to C; and about the Centre F, with the Distance F D, describe a Circle D KL t; and about the Centre G, + 3 Peffo with the Distance GH, describe another Circle KL H, and join KF, KG. I say, 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 shall be equal to FDI: But F D is equal to A ; therefore F K is also equal to A. Again, because the Point G is the Centre of the Circle L K H, GK will be f equal to GH: But G H is equal to C; 1 Def. 15. therefore Thall G K be also equal to C: But F G is likewise equal to B; and consequently the three Right Lines KF, FG, KG, are equal to the three Right Lines A, B, C. Wherefore, ibe Triangle KFG is made of three Right Lines KF, FG, K G, equal to the three given Lines A, B, C, which was to be done. C3 PRO. PROPOSITION XXIII. PROBLEM. in it, to make a Right-lin'd Angle equal to a Right-lin?d Angle given. 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. Assume the Points D and E at Pleasure in the Lines CD, CE, and draw D E; then, of three Right • 22 of this. Lines equal to CD, DE, E C, make * a Triangle AFG, so that A F be equal to CD, AG to CE, and F G to DE. Then because the two Sides DC, CE, are equal to the two Sides F A, AG, each to each, and the Base DE equal to the Base F G; the Angle DCE shall + 8 of tbis. be + equal to the Angle F A G. Therefore, the Right lin'd Angle F A G is made at the given Point A in the given Right Line A B, equal to the given Right-lin'd Angle D CE; which was to be done. PROPOSITION XXIV. THEOREM. to two Sides of the other, each to each, and Ibe will be greater than the Base of the other. LET there be two Triangles A B C, DEF, hava ing two Sides A B, AC, equal to the two Sides DE, Ď F, each to each, viz. the Side A B equal to the Side DE, and the Side A C equal to DF; and let the Angle B A C be greater than the Angle EDF. I say, the Base B C is greater than the Base E E. For For because the Angle BAC is greater than the Angle EDF; make an * Angle EDG at the Point 23 of tbis. D in the Right Line D E, equal to the Angle BAC; and make + DG equal to either AC or DF, and † 3 of tbit. join EG, FG. Cafe 1. When EG falls above E F; then because A B is equal to D Е, 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 Base B C is equal I to the Base E G. 14 of tbit. Again, because D G is equal to D F, the Angle DFG is 7 equal to the Angle DFG; and so the Angle DFG is greater than the Angle EGF: and coniequently the Angle E F G is much greater than the Angle EGF. And because EF G is a Triangle having the Angle EFG greater than the Angle EGF; and the greatest Angle fubtends * the greatest 19 of tbise Side, the Side EG shall be greater than the Side EF. But the Side E G is equal to the Side B C: Whence B C is likewise greater than EF. Cafe 2. When E G falls upon EF; then E G is greater than EF: And consequently B C is greater than E F. Cafe 3. When EG falls below E F; then D G E G; are together greater than D F and E F toge- [ 22 of Ibiza ther, and by taking away the Equals DG, D F, there remain EĠ greater than EĚ* Therefore B C, . Ax. s. which is equal to EG, will be also greater than E F. 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 Base of the one will be greater than the Base of the other ; which was to be demonstrated. If twó Triangles bave two sides of the one equal to two sides of the other, each to each, and the Base af the one greater than the Base of the other ; tbey shall also bave the Angles contained by the equal Sides, the one greater than the otber. LET there be two Triangles ABC, DE F, 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 A C to the Side DF; but the Base BC greater than the Base EF; I say, the Angle BAC is also greater than the Angle E B F. For if it be not greater, it will be either equal or less. But the Angle B A C is not equal to the Angle • 4 of ibis. EDF; for if it was, the Base BC would be * equal to the Base EF; but it is not: therefore the Angle BAC is not equal to the Angle EDF; neither will 24 of this. it be lesser; for if it should, the Base B C would be + less than the Base E F; but it is not. Therefore the Angle BAC is not less than the Angle EDF; but it has likewise been proved not to be equal to it. Where. fore the Angle BAC is necessarily greater than the Angle E DF. If, therefore, two Triangles have two Sides of the one equal to two sides of the other, each to each, and the Base of the one greater than the Base of the other ; they mall also have the Angles contained by the equal Sides, the one greater than the other ; which was to be demonstrated. PRO |