57. That which hath only the oppofite Sides equal and the Angles oblique, is called a Romboides, as D; and may be conceived as an inclined Rectangle. A B C D D 58. When none of the Sides are parallel to another, then the quadrilateral Figure is called a Trapezium. 59. Every other right-lined Figure, that has more Sides than four is in general called a Polygon. And Figures are called by particular Names according to the number of their Sides, viz. One of five Sides is called a Pentagon, of fix a Hexagon, of feven a Heptagon, and so on. When the Sides forming the Polygon are equal to one another, the Figure is called a regular Figure or Polygon. 60. In any Triangle ABC, one of it's Legs, as BC, being produced towards D, the external Angle ACD is equal to both the internal oppofite ones taken together, viz. to ABC and BAC. In order to prove this, through C draw CE parallel to AB; then fince CE is parallel to AB and A C croffeth them, the Angle ECD is equal to ABC (by the 37th) and the Angle ACE equal to CAB (by the 36th) therefore the Angles ECD and ECA are equal to the Angles ABC and CAB; but the Angles ECD and ECA are together equal to the Angle A CD; therefore the Angle ACD is equal to both the Angles ABC and C A B taken together. A E Б D C 61. In any Triangle ABC all the three Angles taken together are equal to two right Angles. To prove this you must produce B C, one of it's Legs, to any distance, fuppofe to D; then by the laft Propofition, the external Angle, ACD, is equal to the Sum of the two internal oppofite ones CAB and ABC; to both add the Angle ACB, then the Sum of the Angles ACD and ACB will be equal to the Sum of the Angles CAB and CBA and A CB. But the Sum of the Angles ACD and ACB, is equal to two right ones (by the 32d) therefore the Sum of the three Angles CAB and CBA and ACB, is equal to two right Angles; that is, the Sum of the three Angles of any Triangle ACB is equal to two right Angles. Cor. 1. Hence in any Triangle given, if one of it's Angles be known, the Sum of the other two is also known; for fince by the laft, the Sum of all the three is equal to two right Angles, or a Semicircle, it is plain, that taking any one of them from a Semicircle or 180 Degrees, the Remainder will be the Sum of the other two. Thus (in the former Triangle ABC) if the Angle ABC be 40 Degrees, by taking 40 from 180 we have 140 Degrees; which is the Sum of the two Angles BAC, ACB, the converse of this is also plain, viz. The the Sum of any two Angles of a Triangle being given, the other Angle is also known by taking that Sum from 180 Degrees. 2. In any right angled Triangle, the two acute Angles muft juft make up a right one between them; confequently, any one of the oblique Angles being given we may find the other by fubtracting the given one from 90 Degrees, which is the Sum of both. 62. If in any two Triangles, A B C, DEF, two Legs of the one, viz. A qual to two Legs in D the other, viz. to DE and DF, each to each respectively, i. e. A B to DE and AC to DF; and if the Angles included between the equal Legs be equal, viz. the Angle BAC equal to the Angle EDF; then I fay, that the remaining Leg of the one fhall be equal to the remaining Leg of the other, viz. BC to EF; and the Angles oppofite to equal Legs fhall be equal, viz. ABC equal to DEF (being oppofite to the equal Legs A ̊C, and D F) alfo ACB equal to DFE (which are oppofite to the equal Legs AB and DE) for if the Triangle ABC be fuppofed to be lifted up and put upon the Triangle DEF, and the point A on the point D; it is plain fince B A and DE are of equal length, the point E will fall upon the point B; and fince the Angles BAC, EDF are equal, the Line AC will fall upon the Line DF, and they being of equal length, the Point C will fall upon the Point F, and fo the Line BC will exactly agree with the Line EF, fo the Triangle ABC will in all refpects be exactly equal to the Triangle DEF; and the Angle ABC will be equal to the Angle DEF, alfo the Angle ACB will be equal to the Angle DFE. 63. Any Angle, as BAD, at the Circumference of a Circle BADE, is but half the Angle BCD at the Center ftanding on the fame Arch BED. To demonftrate this, draw through A and the Center C, the right Line ACE, then the Angle ECD is B A D E equal equal to both the Angles DAC and ADC (by the 60th); but fince AC and CD are equal (being two Radii of the fame Circle) it is plain the Angles fubtended by them must be equal also, i. e. the Angle CAD equal to the Angle CDA, therefore the Sum of them is double any one of them, i. e. DAC and A D C is double of CAD, and therefore ECD is alfo double of DAC; the fame way it may be proved, that ECB is double of CA B, and therefore the Angle BCD is double of the Angle BAD, or BAD the half of BCD which was to be proved. Cor. 1. Hence an Angle at the Circumference is measured by half the Arch it fubtends, for the Angle at the Center (ftanding on the fame Arch) is measured by the whole Arc (by the 30th); but fince the Angle at the Center is double that at the Circumference, it is plain the Angle at the Circumference must be measured by only half the Arch it ftands upon. A D E Cor. 2. Hence all Angles, ACB, ADB, AEB, &c. at the circumference of a Circle, ftanding on the fame Chord AB, are equal to one another; B for by the laft Corollary they are all measured by the fame Arc, viz. half the Arc AB which each of them fubtends. on Cor. 3. Hence an Angle in a Segment greater than a Semicircle is lefs than a right Angle; thus if ADB be a Segment, greater than a Semicircle, (fee the laft Figure) then the Arch AB, which it ftands, must be less than a Semicircle, and the half of it less than a Quadrant or a right Angle; but the Angle ADB in the Segment, is measured by the half of AB; therefore it is lefs than a right Angle. Cor. 4. An Angle in a Semicircle is a right Angle. For fince ABD a Semicircle, the Arc AED muft alfo be a Semicircle; but the Angle ABD is measured by half the Arc AED, that is, by half a Se A B micircle or Quadrant; there fore the Angle ABD is a right one. E A B D Cor. 5. Hence an Angle in a Segment less than a Semicircle, as A BD, is greater than a right Angle: for fince the Arch ABD is lefs than a Semicircle, the Arch AED must be greater than a Semi-. circle, and fo it's half greater than a Quadrant, i. e. than the measure of a right Angle; therefore the Angle ABD, which is measured by half the Arch AED, is greater than a right Angle. E 64. If from the Center C of the Circle A B E, there be let fall the Perpendicular CD on the Chord A B, then that Perpendicular will bifect the Chord AB in the Point D. To demonftrate this, draw from the Center to the Extremities of the Chord the two Lines CA, CB; then fince the Lines CA and CB are equal, the Angles CAB, CB A, which they fubtend must be equal alfo; but the Perpendicular * CD divides the Triangle ACB into two right angled Triangles ACD and CDB, in which the Sum of the Angles ACD and CAD in the one, is equal to the Sum of the Angles DCB and CBD in the other, each E A being equal to a right Angle, (by Cor. 2. of Art. 61.) but CAD is equal to CB D, B therefore ACD is equal to BCD. So in the two |