For produce BC to D. Then, because the outward Angle A CD of the 16 of this. Triangle A B C is greater than the inward oppofite Angle ABC; if the common Angle A C B be added, the Angles ACD, ACB, together, will be greater than the Angles ABC, ACB, together: But ACD, +13 of this. A CB, are † equal to two Right Angles. Therefore ABC, BCA, are lefs than two Right Angles. In the fame manner we demonftrate, that the Angles BAC, ACB, as alfo C A B, ABC, are less than two Right Angles. Therefore, two Angles of any Triangle together, bowfoever taken, are lefs than two Right Angles; which was to be demonftrated. L greater Angle, ET ABC be a Triangle, having the Side A C greater than the Side A B. I fay, the Angle ABC is greater than the Angle B C A. For, because AC is greater than AB, AD may ↑ 3 of this, made equal to A B‡, and BD be joined. be Then, because A D B is an outward Angle of the * 16 of this. Triangle B DC, it will be* greater than the inward ts of this. oppofite Angle DCB. But ADB is equal to ABD; because the Side A B is equal to the Side AD. Therefore the Angle ABD is likewife greater than the Angle ACB; and confequently ABC fhall be much greater than A CB. Wherefore, the greater Side of every Triangle fubtends the greater Angle which was to be demonftrated. PRO PROPOSITION XIX. THEORE M. The greater Angle of every Triangle fubtends the greater Side.. ET ABC be a Triangle, having the Angle A B C greater than the Angle BCA. I fay, the Side A C is greater than the Side A B. * * For, if it be not greater, AC is either equal to A B, or less than it. It is not equal to it, because then the Angle A B C would be equal to the Angle A CB; 5 of this. but it is not: Therefore AC is not equal to A B. Neither will it be lefs; for then the Angle ABC would be + less than the Angle A CB; but it is not. † 18 of this. Therefore AC is not less than A B. But likewife it has been proved not to be equal to it: Wherefore AC is greater than A B. Therefore, the greater Angle of every Triangle fubtends the greater Side; which was to be demonstrated. PROPOSITION XX. THEOREM. Two Sides of any Triangle, howsoever taken, are togeiber greater than the Third Side. L ET ABC be a Triangle. I fay, two Sides thereof, howsoever taken, are together greater than the third Side; viz. the Sides BA, A C, are greater than the Side BC; and the Sides A B, BC, greater than the Side A C; and the Sides B C, CA, greater than the Side A B. For produce B A to the Point D, so that AD be * equal to A C; and join DC. * 3 of this. Then, because DA is equal to AC, the Angle ADC fhall be equal to the Angle ACD. Butthe† 5 of this. Angle BCD is greater than the Angle ACD. Wherefore the Angle B C D is greater than the Angle ADC; and because DCB is a Triangle, having the Angle B C D greater than the Angle BDC, and the greater C 2 † 19 of this greater Angle fubtends + the greater Side; the Side DB will be greater than the Side BC. But DB 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 AB, BC, together, are greater than the Side CA; and the Sides BC, CA, 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 demonstrated. PROPOSITION XXI. THEORE M. If two Right Lines be drown from the extreme Points of one Side of a Triangle to any Point within the fame, these two Lines shall be less than the other two Sides of the Triangle, but 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 ABC, to the Point D within the fame. I fay, BD, DC, are lefs than B A, A C, the other two Sides of the Triangle, but contain an Angle BDC greater than the Angle B A C. * 20 of this. are + Ax. 4. For produce B D to E. * Then, because two Sides of every Triangle together greater than the third, BA, A E the two Sides of the Triangle ABE, are greater than the Side BE. Now, add EC, which is common, and the Sides BA, A C, will be + greater than BE, EC. Again, because CE, ED, the two Sides of the Tris angle CED, are greater than the Side CD, add DB, which is common, and the Sides CE, EB, will be greater than CD, D B. But it has been proved, that BA, AC, are greater than BE, EC. Wherefore BA, AC, are much greater than BD, DC. Again, 16 of ibis. because the outward Angle of every Triangle is greater than the inward and oppofite one; B DC, the outward Angle of the Triangle CDE, fhall be greater than the Angle CED. For the fame Reason, CED, the outward Angle of the Triangle A BE, is likewife greater |