= A thereof make a an Angle DAE ADC: then a 27.1. will AE and BC be parallel. Q. E. D. If one Side (BC) of any Triangle E (ABC) be continued out; the outward Angle ACD D shall be equal to : the two inward oppofite Angles A, B and the three inward Angles A, B, ACB, of a Triangle, shall be equal to two right Angles. а From C draw a CE parallel to BA; then is a 31. 1. the Angle Ab ACE, and the Angle Bbb 29. 1. ECD: Therefore A+B ACE+ECD dc 2 ax. ACD. Q. E. D. d 19 ax. Again, ACD+ ACB two right Angles :e 13. 1. Therefore A+B+ACBf two right An-f 1 ax. gles. Q. E.D. COROLLARIE S. 1. The three Angles of any Triangle taken together, are equal to the three Angles of any other Triangle taken together. From whence it follows, 2. That if in one Triangle two Angles (taken feverally or together) be equal to two Angles. of another Triangle, (taken feverally or together) then is the remaining Angle of the one, equal to the remaining Angle of the other. In like manner, if two Triangles have one Angle of the one, equal to one of the other; then is the Sum of the remaining Angles of the one Triangle, equal to the Sum of the remaining Angles of the other. 3. If one Angle in a Triangle be a right Angle, the other two are equal to a right one. Like Likewife that Angle in a Triangle, which is equal to the other two, is itself a right Angle. 4. In an Ifofceles Triangle, if the Angle made by the equal Sides is a right one, the other two upon the Bafe are each of them half a right Angle. 5. An Angle of an equilateral Triangle is two thirds of a right Angle: for of two right Angles is equal to of one. SCHOLIU M. By the help of this Propofition, you may know how many right Angles the inward and outward Angles of a right-lined Figure make; as will appear by these two following Theorems. THEOR. I. All the Angles of a right-lined Figure do together make twice as many right Angles, abating four, as there are Sides of the Figure. ADO From any Point within the Figure draw right Lines to all the Angles of the Figure, which fhall refolve the Figure into as many Triangies as there are Sides of the Figure. Since therefore in every Triangle the Sum of all the Angles is two right Angles, all the Angles of the Triangles taken together will make up twice as many right Angles as there are Sides. But the Angles about the faid Point within the Figure make four right Angles therefore if from the the Angles of all the Triangles, you take away the Angles which are about the faid Point, the remaining Angles, which make up the Angles of the Figure, will make twice as many right Angles, abating four, as there are Sides of the Figure. QE. D. CORO L.. Hence all right-lined Figures of the fame Species or Kind have the Sums of their Angles equal. THEOR. II. All the outward Angles of any right-lined Figure, taken together, make four right Angles. For every inward Angle of a Figure, with the outward Angle of the fame, make two right Angles; therefore all the inward Angles together, with all the outward Angles, make twice as many right Angles as there are Sides of the Figure. But (as it was juft now fhewn) all the inward Angles, together with four right Angles, make twice as many right Angles as there are Sides of the Figure; therefore the outward Angles are equal to four right Angles. QE.D. COROL. The Sum of the outward Angles of any rightlined Figure is equal to the Sum of the outward Angles of any other right-lined Figure.. Draw a Line from C to B: because AB and CD are parallel, the Ang. ABC BCD: and (by Hypothefis) ABCD, and the Side CB common; therefore AC BD, and the Ang. ACB DBC: Whence alfo AC, BD are parallel. C The oppofite Sides AB, CD, and AC, BD, of a Parallelo C gram, as ABDC, are equal each to the other; as also the oppofite Angles A,D; and ABD, h i 27. I. ACD: and the Diameter BC bifects the fame. d с d e Because AB, CD are parallel, therefore the the Ang. ABC = BCD. Also because AC, ED are parallel, the Angle ACB fhall be "= CBD; therefore the whole Ang. ACDf=ABD. After the fame manner, AD. Moreover, because the Angles ABC, ACB are adjacent to the common Side CB, and are equal to BCD, CBD; therefore AC = BD, and AB="CD; and fo the Triangle ABC= CBD. Q.E.D. SCHOLIUM. Every quadrilateral or four-fided Figure ABDC, having the oppofite Sides equal, is a Parallelogram. For (by 8. 1.)the Ang. ABC BCD; therefore AB, CD are " parallel. In like manner the Ang. BCA is CBD; therefore AC, BD" are 35 def. alfo parallel whence ABDC is a Parallelogram. Q. E. D. i Hence CD, to a given through an affigned Point C. In the Line AB take any Point, as E; and about the Centers E and C with any Distance draw two equal Circles EF, CD: alfo at the Center F, with the Distance EC draw a Circle FD, which fhall cut the former Circle (CD) in the Point D: then fhall the Line drawn (CD) be parallel to AB; for we demonftrate as before that CEFD is a Parallelogram. Parallelograms (BCDA, BCFE) conftituted upon the fame Bafe BC, and between the fame Parallels AF, BC, are equal the one to the other. C a 29. I. 4. 1. For AD BC= EF, add DE common 34. I. to both; then AE= DF but alio AB a b 11 2 ax. DC, and the Ang. A CDE: therefore the Triangle ABEDCF. Take away the Triangle DGE common to both, then the Trapezium ABGD EGCF: add the Triangle BGC common to both; and then the Parallelo- f 2 ax. gram ABCDEBCF. Q. E. D. e 3 ax. |