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adjacent to those sides in the A QRS = to the <s corresponding to them in the Triangle grs, all the reft, and also the Trianglès them. selves, will be equal.

For if the side or be put upon the side QR, they will agree (by Maxim 7.) but b.cause the < R= <r, and <S=<s; the fide Rg will fall upon fide RQ, and as upon QS, .: the point q will fall upon the point; (for if it fall out of Q, the Lines r q, qs do not fall upon the Lines R, QS) therefore they are equal (by the seventh Maxim) Q.ED.


Fig.37. In an Isoceles Triangle the Angle at the Base opposite to the equal Legs, are equal.

Let the Triangle A CB be supposed two Triangles, and the Situation of one con Vers to that of the other, as bca: Because in the two Triangles ACB and b ca, the side AC is = the fide b c, and the side CB is the side ab, and the Angle C is = the Angle c, therefore the Angle at the Base A =< b by the first, which was to be Demonstrated for the Angle, B and b are the fame.



Fig. 38. If two Triangles have each side in one equal to its corresponding fide in the other (tbar is, ac = ef, cb=fi, and ab = ei). they will allo have = Angles opposite to those = fides (that is c = f, a=e,and b=i).

Let the line ab be put upon the line ei. Then the point c will either fall in f, or it will not. If it falls in f, the whole Triangles gree, and therefore all the Angles are equal by the seventh Maxim.

Fig 39. If c falls out of f, draw the line fc: Because by Hypothesis the sides ef, and a c are equal, the

Angle efc inust be equal to the Angle ect; by the second Propofition therefore Angle ife will be greater than Angle ecf, and Angle ifc wil be much greater than An.

Again, by supposition, because if = bc; <ifc, will be = to sicf. Therefore ifcis both much greater than, and equal ta cf, which is impossible, and theretore a cannot fall out of f.

gle icf.


Fig.40.One right line CDalling on another AB, makes the Contiguous Angles = to two right Angles. Let the line CD be perpendicular to AB, then the Angles ADC and CDB

will be right pr.5.Def. But if the line fall obliquely as ED, raise the perpendicular CD), then, because the unequal. Angles ADE and EDB occupy the same place which the two right ones. ADC and CDB did, therefore they agree to two right Angles, and consequently to them (by 7 Ax.) Q.E.D.


Fig.41. If two right lines (BC, and FL) cut one che other in any point (A) the opposite Angles at the Vertex (A) will be equal; that is, the Angle LAB is = to CAF' Becanse BA, stands upon the right line LF, LAB, and FAB will be equal to two right (by the fourth Theorem.) And because FA Itands upon BC, the Angles' FAC, and FAB, will be equal to two right (by the fame) therefore the two Angles LAB, and FAB taken together will be és tal to the two Angles CAF, and FAB, Dicen together. But the common Angle FAB being taken away, there remains the Angle LAB = CAF (by the third Max.) Q.E.D.


Fig 42. If a right line GO cut the two parallel right lines (AB CF,) first the alternate <(RLO, QOL, and BLO,COL.) are equals. Secondly, The external Angle (GLB) is equal

to the internal one (LOF,) as also(< GLR= <LOC.) Thirdly, The two internal Angles towards the fame parts together ALO COL= to two right Angles : Also the 58 BLO, FOL together = to two right Angles,

Demon. of the first part Draw LQ and RO perpendicular toCF from the points L ando and they will be also perpendicular to AB(by tbe 1 2th Maxim,) now in the as ROL, LOQ, the side OL., is common to both; RO=LQ (by the 8th Def.)and .::<LOR=<LOQ (by the first Prop.) also <ROL=<QL0.5 BLO =<COL (by the second Max.) which is the First Part,

Second Part. Angle LOQ =<ROL (by tbe Firsi Part,) and <ŘLO= CGLB (by the stb Prop.): <LOQ=< GLB (by the firf Max.) after the same manner may <LOC be

be proved =tc< GLR which is the second part,

The 3d. Part, Angle GLB =<RLO (by the 5th Prop.) <COL was proved => RLG (in the 2d. Part.). but < RLG +<GLB = two right <s (by the 4th Prop.)..< COL +<RLO = two right Angles, as allo s BLO + <FOL Q. E. D.


In any Triangle abc, the three Angles taken together, are equal to two right ones.

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