than the Angle BAC; but the Angle BDC has been proved to be greater than the Angle CEB. 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; these 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.

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 DE terminated at D, but infinite towards E; and take *DF * 3 of this equal to A, FG equal to B, and GH equal to C; and about the Center F, with the Distance FD, defcribe a Circle DKL † ; and about the Center G, with † 3 Poff the Distance GH, 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 Center of the Circle DK; FK fhall be equal to FD; but FD is equal to A; therefore F K is also equal to A. Again, because the Point G is the Center of the Circle LKH, GK will be ‡‡ Def. 15: equal to GH; but GH is equal to C: Therefore fhall GK be alfo equal to C; but FG 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 KFG is made of three Right Lines KF, FG, GK, equal to the three given Lines A, B, C, which was to be done.

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.

ET the given Right Line be AB, and the Point given therein A, and the given Right-lined Angle DCE. It is required to make a Right-lined Angle at the given Point A, with the given Right Line AB, equal to the given Right-lined Angle DCE.

Affume the Points D and E at Pleasure in the Lines CD, CE, and draw DE; then, of three Right Lines 22 of this. equal to CD, DE, EC, make* a Triangle AFG, fo that AF be equal to CD, AG to CE, and FG to DE.

Then because the two Sides DC, CE, are equal to the two Sides FA, AG, each to each, and the Base DE equal to the Base F G; the Angle DCE fhall be + 8 of this. + equal to the Angle FAG. Therefore the Rightlined Angle FAG is made, at the given Point A, in the given Line AB, equal to the given Right-lined 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 Bafe of the one will be greater than the Bafe of the other.

ET there be two Triangles ABC, DEF, having two Sides AB, AC, equal to the two Sides DE, DF, each to each, viz. the Side AB equal to the Side DE, and the Side AC equal to DF; and let the Angle BAC be greater than the Angle EDF. I fay, the Base BC is greater than the Bafe EF.

*

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

and make + DG equal to either AC or DF, and † 3 of this. join EF, FG.

Now because AB is equal to DE, 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 Bafe BC is equal † to 14 of this. the Bafe EG. Again, becaufe DG is equal to DF, the Angle DFG is equal to the Angle DGF; and ↓ 5 of this. fo the Angle DFG is greater than the Angle EGF: And confequently the Angle EFG is much greater than the Angle ÉGF. And because EFG is a Triangle, having the Angle EFG greater than the Angle EGF; and the greateft Side fubtends the greatest ‡ 19 of this. Angle, 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 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 Bafe of the other; which was to be demonstrated.

PROPOSITION XXV.

THEOREM.

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 fhall alfo bave the Angles, contained under the equal Sides, the one greater than the other.

ET there be two Triangles ABC, DEF, having two Sides AB, AC, each equal to two Sides DE, DF, viz. the Side AB 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 also greater than the Angle EDF.

For if it be not greater, it will be either equal or lefs. But the Angle BAC is not equal to the Angle

C 3

EDF;

* 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 it t24 of this. be leffer; for if it fhould, the Bafe BC would be+ lefs. than the Base EF; 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. 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 alfo have the Angles, contained under the equal Sides, the one greater than the other ; which was to be demonftrated.

PROPOSITION XXVI.

THEOREM.

If two Triangles have 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 shall 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.

ET there be two Triangles A B C, DEF, hav

ing two Angles ABC, BCA 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 BCA equal to the Angle EFD. And let one Side of the one be equal to one Side of the other, which firft let be the Side lying between the equal Angles, viz. the Side BC equal to the Side EF. 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 AB 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

For if the Side A B be not equal to the Side DE, one of them will be the greater, which let be AB, make GB equal to DE, and join G C.

Then because BG is equal to DE, and BC to EF, the two Sides G B, BC, are equal to the two Sides DE, EF, each to each; and the Angle GBC equal to the Angle DEF. The Base GC is * equal to the* 4 of this. Bafe DF, 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 DFE. But the Angle DFE, by the Hypothefis, is equal to the Angle BCA; and fo the Angle BCG is likewise equal to the Angle BCA, the less to the greater, which cannot be. Therefore A B is not unequal to DE, and confequently is equal to it. And fo the two Sides AB, BC, are each equal to the two Sides DE, EF, and the Angle ABC equal to the Angle DEF: And confequently the Base AC * is equal to the Base DF, and the remaining Angle BAC equal to the remaining Angle EDF.

Secondly, Let the Sides that are fubtended by the equal Angles be equal, as AB equal to DE. I fay, the remaining Sides of the one Triangle, are equal to the remaining Sides of the other, viz. AC to DF, and BC to EF; and also the remaining Angle BAC, to the remaining Angle EDF.

For if BC be unequal to EF, one of them is the greater, which let be BC, if poffible, and make BH equal to EF, and join A H.

*

Now because B H is equal to EF, and AB to DE, the two Sides AB, BH, are equal to the two Sides DE, EF, each to each, and they contain equal Angles: Therefore the Bafe AH is equal to the Bafe DF; and the Triangle ABH fhall be equal to the Triangle DEF, and the remaining Angles equal to the remaining Angles, each to each, which fubtend the equal Sides: And fo the Angle BHA is equal to the Angle EF D. But EFD is + equal to the Angle † From the Hyp. BCA; and confequently the Angle BHA is equal to the Angle BCA: Therefore the outward Angle BHA of the Triangle AHC, is equal to the inward and oppofite Angle BCA; which is impoffible: 16 of this Whence BC is not unequal to EF, therefore it is Ç 4 equal