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ward Angle is greater iban either of ihe inward
oppohte Angles. LET A B C be a Triangle, and one of its Sides
B C be produced'to D. I say, the outward Angle A CD is greater than either of the inward Angles CBA, or BA C.
For bisect A C in E*, and join BE, which pro.. 10 of tbis, duce to F, and make E F equal to B E +. Moreover,
+ 3 of bisa join F C, and produce A C to G.
Then, because AE is equal to E C, and B E to EF, the two Sides A E, E B, are equal to the two Sides CE, E F, each to each, and the Angle A EB + equal to the Angle FEC; for they are opposite + 15 of ibis, Angles. Therefore the Bale A B is f equal to the 14 of sbis. Bale FC; and the Triangle A E B equal to the Triangle FEC; and the remaining Angles of the one equal to the remaining Angles of the other, each to each, subtending the equal Sides. Wherefore the Angle B A E is equal to the Angle ECF; but the Angle E C D is greater than the Angle ECF; therefore the Angle A C D is greater than the Angle B A E. After the same manner, if the Right Line B C be bisected, we demonstrate that the Angle BCG, and consequently its equal, the Angle A CD*, is greater 15 of tbis. than the Angle A BC. Therefore, one side of any Triangle being produced, the outward Angle is greater than either of the inward opposite Angles; which was to be demonstrated.
taken, are less than two Right Angles.
of it together, howsoever taken, are less than two Right Angles.
For produce B C to D.
Then, because the outward Angle ACD of the • 16 of tbis. Triangle A B C is greater * than the inward opposite
Angle A BC; if the common Angle A C B be added, the Angles A C'D, ACB, together, will be greater
than the Angles ABC, ACB, together: But ACD, † 13 of tbis. ACB, are + equal to two Right Angles. Therefore
ABC, BCA, are less than two Right Angles. In the same manner we domonftrate, that the Angles BAC, ACB, as also C A B, ABC, are less than two Right Angles. Therefore, two Angles of any Triangle together, howsoever taken, are less than tuo Right Angles; which was to be demonstrated,
The greater Side of every Triangle subtends the
greater Angle. ET AB C be a Triangle, having the Side A C
greater than the Side A B. I say the Angle ABC is creater than the Angle B CA.
For, because AC is greater than A B, A D may be 1 3 of tbit. made equal to A B I, and B D be joined.
Then, because A D B is an outward Angle of the • 16 of tbis. Triangle BDC, it will be * greater than the inward #softbis. oppofite Angle DCB. But ADB is † equal to
ABD; because the Side A B is equal to the Side A D. Therefore the Angle A B D is likewise greater than the Angle A CB; and confequently ABC thall be much greater than ACB. Wherefore, the greater Side of every Triangle subtends the greater Angle; which was to be demonstrated,
ET A B C be a Triangle, having the Angle
A B C greater than the Angle BCĂ. I say, the Side A C is greater than the Side A B.
For, if it be not greater, AC is either equal to AB, or less than it. It is not equal to it, because then the Angle ABC would be equal * to the Angle ACB:
5 of this. but it is not: Therefore A C is not equal to A B. Neither will it be less; for then the Angle ABC would be + less than the Angle ACB; but it is not. + 18 of this. Therefore A Cis not less than AB. But likewise it has been proved not to be equal to it: Wherefore AC is greater than A B. Therefore, the greater Angle of every Triangle subtends the greater Side : which was to be demonstrated.
together greater Than the Tbird Side.
thereof, howsoever taken, are together greater. than the third Side ; viz. the Sides B A, AC, greater than the Side B C; and the Sides A B, BC, greater than the Side A C; and the sides BC, CA, greater than the Side A B.
For produce B A to the Point D, so that A D be equal to AC; and join D C.
* 3 of this. Then, because D A is equal to A C, the Angle ADC shall be equal + to the Angle ACD. But + s of tbisa the Angle BCD is greater than the Angle ACD. Wherefore the Angle BCD is greater than the Angle ADC; and because D C B is a Triangle, having the Angle B C D greater than the Angle BDC; and the C2)