DEMONSTRATION. Fig.43. Let d be drawn parallel to the Base ac, then the <a will < dba (by the 6th Prop.) and <c, = < ebc (by the 6th. Prop.) butd ba - abc + < cbe Angles (by the 4th Prop.). < a abctwo right Q. E. D. to two right <c+< CORROL ARTI. Hence by Having one of the acute Angles in any Rectangled ▲, you may find the cther by taking the Angle out of 90 Degrees, and that which remains will be the Angle required; fo that one Angle is the Complement of the other to 90 Degrees: If your Triangle be obtufe or acute, any two of the Angles being known, the 3d is known, by taking the Sum of the two Angles known, from 180 deg. and the remainder is the other Angle. CORROL ART. II. Fig. 1.If any fide of a Triangle be produced, the external Angle o will be equal to the two Internal and oppofite Angles b and e. For the Angles b and e together with d, are equal to two right ones (by the Preced.) and fo alfo are the Angles O and d,. (by the 4th) where wherefore O muft be equal to b added to e, because together with d, it makes two right ones, as they do. Q. E.D. CORROL ART. III. Fig. 45. The internal Angles A, B, C, D, E, of any Polygon ABCDE are equal to twice as many right Angles as it has fides, except four. For every Polygon may be divided into as many Triangles,except two, as it has fides, by lines drawn from any Angle D to all the reft, except the adjoining ones A and E; but each of thofe Triangles contain three Angles two right ones (by the 7th) therefore all the Angles of any Polygon are equal to twice as many right Angles, except four. CORROL ART. IV. Fig.46. In any Polygon A B CD E, the fides being produced will make all the external Argles a, b, c, d, e, equal to four right ones: For each internal< Awith its external< a, is equal to two right Angles (by the 4th i)confequently all the internal and external Angles are equal to twice as many right Angles as the Figure has fides. But all its internal Angles are equal to twice as many right Angles except 4 as it has fides (by the laft Cor.) therefore all the external Angles taken tegether, are equal to four right ones. CORRO LART V. From the third Corrolary we may de duce a Method to examine whether the Angles of a Field be taken right or not, viz. Multiply the number of Sides thereof by 2; fubftract 4 from that product, which remainder multiply'd by 90, will give the Sum of all the Angles, which Sum must be equal to the Sum of the Angles taken in the Field if rightly obferv'd. Fig. 46 For Example: Let ABCDE be a Field bounded with 5 fides, the Sum of the Angles whereof is 540° which is equal to the Number of fides (5) multiply'd by 2, and that Product leffen'd by 4, multiply'd by 90, therefore it is very probable that thofe Angles were rightly obferv'd. THEOREM VIII. Fig.47.The oppositeSides and Angles of any Parallelogram are equal, and it is divided into two equal parts by (a Line drawn from one to its oppofite one which is called) a Diagonal Line. Because O B is parallel to DC (by Def. 14.) the < OBD = < BDC (by Prop. 6) alfo because DO and BC are parallel < CBI) = < ODC (by the fame). the whole < OBC = ODC after the fame manner <O may be C thewa fhewn to be equal to < C. But fince we have fhewn that thefe Triangles Q and R which have one fide D B common, and two Angles adjacent to D B in one, equal to the two correfponding Angles in the other, therefore thofe Triangles and all their parts,are qual by the Scholium to the Prop.1. SCHOLIUM. Fig. 48. For the Triangle AEF and GFD, having the Alternate Angles EAF and FDG, and AEF and FGD, and AF and FD qual they are equal, (by the 1ft,) and fince the Trapezium BEFD, with the Triangle AEF, that is to fay, the Triangle ADB, is half the Parallelogram, (by the laft) the fame Trapezium BEFD, with the Triangle FGD, will be half the fame; therefore the line EG divides it into into two qual parts. THEOREM IX. Fig.49.Parallelograms having the fame Base, and being between the fame Parallels, are equal. For AB = EC &c. (by the 8th) BAFA (ECG by the 1ft) from both of which take away the common ▲ DEF, and there will remain the Trapezium AEDB = Trapezium DCFG, to each of which add the BDC and then the Parallelogram ABEC = Parallelogram FBCG per 2d Mex. THEOREM X. Fig. 50. Triangles on the fame Bafe AB, and being between the fame Parallels CF and AB are always equal. The Triangle ABC, is equal to the Triangle ADB. Draw the line LB parallel to AC, and the line BF parallel to AD, then there will be made two Parallelograms, A CDB; and ADFB, which being between the fame parallels and on the fame Bafe, will be equal to one another (by the laff;) but the Triangle ACB is the half of the Parallelogra A BCD, and the Triangle DBA is the half of a Parallelogram ABDF, (by the 8th,) Therefore the Triangles A CB and A DB, (must be equal, by the 6th Max. CORROLART. Triangles having the fame or equal Bifes to Parallelograms, and between the fame parallels, are juft half thofe Parallelograms. THEOREM XI. Fig. 51. The Complements of a Parallelogram are equal. In the Parallelogram ABDC the Complements AFEH and EGDI are Equal G 2 DEMON |