† 19 of this. greater Angle fubtends † 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, AC, together, are greater than the Side B C. In the fame manner we demonftrate, that the Sides A B, B C, together, are greater than the Side CA; and the Sides BC, C A, together, are greater than the Side A B. Therefore, two Sides of any Triangle, bowfoever taken, are together greater than the third Side; which was to be demonftrated.

PROPOSITION XXI.

THEOREM.

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 lefs than the other two Sides of the Triangle, but contain a greater Angle.

FO

I

OR let two Right Lines B D, DC, be drawn from the Extremes B, C, of the Side B C of the Triangle A B C, to the Point D within the fame. fay, BD, DC, are less than B A, A C, the other two Sides of the Triangle, but contain an Angle BDC 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 two Sides of the Triangle A B E, are greater than the Side B E. Now, add E C, which is common, and the Sides BA, A C, will be + greater than B E, E C.

+ Ax. 4.

116 of this.

Again, becaufe CE, ED, the two Sides of the Triangle C E D, are greater than the Side C D, add DB, which is common, and the Sides C E, E B, will be greater than C D, D B. But it has been proved, that BA, A C, are greater than B E, E C. Wherefore BA, A C, are much greater than B D, DC, Again, because the outward Angle of every Triangle is greater than the inward and oppofite one; BDC, the outward Angle of the Triangle CD E, fhall be greater than the Angle CED. For the fame Reafon, ČED, the cutward Angle of the Triangle ABE, is likewife

greater

greater than the Angle B AC; but the Angle BDC has been proved to be greater than the Angle CED. Wherefore the Angle B D C 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 lefs than the other two Sides of the Triangle, but contain a greater Angle, which was to be demonftrated.

PROPOSITION XXII.

PROBLEM.

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.

L

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 E, terminated at D, but infinite towards E; and take * DF * 3 of this. equal to A, F G, equal to B, and G H equal to C; and about the Centre F, with the Distance F D, defcribe a Circle D KL+; and about the Centre G, † 3 Poft. with the Distance G H, defcribe another Circle KL H, and join K F, KG. I fay, the Triangle KF G, 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 F D : But F D is equal to A; therefore F K is alfo equal to A. Again, because the Point G is the Centre of the Circle L KH, GK will be equal to G H: But G H is equal to C; Def. 15. therefore shall G K be alfo equal to C: But F G is likewife equal to B; and confequently the three Right Lines KF, FG, K G, are equal to the three Right Lines A, B, C. Wherefore, the Triangle KFG is made of three Right Lines KF, FG, K Ğ, equal to the three given Lines A, B, C, which was to be done. PRO.

C 3

22 of this.

PROPOSITION XXIII.

PROBLEM.

With a given Right Line, and at a given Point in it, to make a Right-lin'd Angle equal to a Right-lin'd Angle given.

LET 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 D C E.

Affume the Points D and E at Pleasure in the Lines CD, CE, and draw DE; then, of three Right Lines equal to CD, DE, EC, make a Triangle AFG, fo that A F be equal to CD, AG to CE, and F G to D E.

Then because the two Sides DC, CE, are equal to the two Sides F A, A G, each to each, and the Bafe, DE equal to the Base F G; the Angle D C E fhall + of this. be + equal to the Angle FA G. Therefore, the Rightlin'd Angle FA G is made at the given Point A in the given Right Line A B, equal to the given Right-lin'd Angle DCE; which was to be done.

PROPOSITION XXIV.

THEOREM.

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 Bafe of the other.

LET

ABC, 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 A C equal to D F; and let the Angle B A C be greater than the Angle EDF. I fay, the Bafe BC is greater than the Bafe E E.

For

*

For because the Angle B A C is greater than the Angle EDF; make an Angle EDG at the Point 23 of this. D in the Right Line D E, equal to the Angle BAC;

and make + DG equal to either AC or D F, and † 3 of thit, join EG, F G.

14 of this.

Cafe 1. When E G falls above E F; then because A B is equal to D E, and AC to DG, the two Sides BA, AC, are each equal to the two Sides ED, DG, and the Angle BAC equal to the Angle EDG: Therefore the Base B C is equal to the Bafe E G. Again, becaufe DG is equal to D F, the Angle DFG ist equal to the Angle DFG; and fo 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 EFG is a Triangle having the Angle EFG greater than the Angle EGF; and the greateft Angle fubtends the greatest 19 of tbit. Side, the Side EG fhall be greater than the Side EF. But the Side E G is equal to the Side BC: Whence B C is likewife greater than EF.

Cafe 2. When E G falls upon EF; then E G is greater than EF: And confequently BC is greater than E F.

Cafe 3. When EG falls below EF; then DG, EG, are together greater than DF and E F toge, † 21 of thisa ther, and by taking away the Equals DG, D F, there remain E G greater than EF*. Therefore B C, * Ax. 5. 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 correfpondent Angle of the other; then the Bafe of the one will be greater than the Base of the other which was to be demonftrated.

PROPOSITION XXV.

THEORE M..

If two Triangles have two 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 other; they shall alfo have the Angles contained by the equal Sides, the one greater than the other.

ET there be two Triangles A B C, 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 D F ; but the Base B C greater than the Bafe EF; I fay, the Angle B A C is also greater than the Angle E D F.

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 E F; but it is not: therefore the Angle BAC is not equal to the Angle EDF; neither will 24 of this. it be leffer; for if it should, the Bafe BC would be + lefs than the Bafe E F; but it is not. Therefore the Angle BA C is not less 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, two Triangles have two 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 other; they fhall also have the Angles contained by the equal Sides, the one greater than the other; which was to be demonftrated.

PRO