Lib. I. A T a given Point in a right Line (as B) to make Fig. 40 an Angle equal to a given one (A). 1 First of all let CF be drawn at a venture, cutting the Sides of the given Angle A. Then in the given right Line from B take B L equal to A F. Then from the Centre B describe a Circle with the Interval AC; afterwards another from the Centre L with the Interval FC, which may cut the former in O. Then from O unto B and L having drawn right Lines, the Angle LBO will be equal to the given one A. For by the Construction the Triangles are Equilateral to one another. Therefore by the 8th of this Book the Angles B and A are equal. Scholium. IT seems meet for the Sake of beginners to propound some things here which are necessary for Practice a bout Angles. The Measure of an Angle is the Arch of a Circle, Fig. 41. which is described from A, the Top of the Angle as the Centre. Therefore look how many Degrees the Arch BC which is intercepted between the Legs of the Angle BAC shall contain, of so many Degrees the Angle BAC shall be faid to be. And so because BF a quarter of the Circumference, contains 90 Degrees, and mea fures the right Angle BA F, a right Angle shall be faid to be of 90 Degrees. In like manner, because half the Circumference, which is divided into 180 Degrees, measures two right Angles, and the whole Circumference, which is divided into 360 Degrees, measures four right Angles; two right Angles shall be faid to make 180 Degrees, and four, 360 Degrees. These things being premised, the Practice about Angles is as follows. 1. At B a given Point in a right Line to make an Fig. 42 Angle equal to the given one A. - From A the Top of the given Angle as the Centre describe betwixt the Sides the Arch CF. Then from B Fig. 43. Fig. 43. Fig. 44. the given Point as the Centre describe with the same Interval the Arch LZ; from which take off LO equal to C F. Thro' B and O draw a right Line; LBO shall be equal to the given A. 2. To examine the Degrees of the given Angle O P Q. This is done very easily by any Semicircle or Protractor, which is divided into 180 Degrees. For put the Centre of the Semicircle upon P the Top of the Angle, and the Radius of the Semicircle P L upon the Side of the Angle PQ; and the Arch LO which is intercepted betwixt the Legs of the Angle will shew of how many Degrees the given Angle is. 3. To frame an Angle containing a given Number of Degrees, as 42. Draw the right Line XQ, in which mark the Point P. Upon P put the Centre of the Semicircle, and its Semidiameter PL upon PQ. From L number 42 Degrees, that is, until you come to O. A right Line drawn from P thro' O, will give the Angle OPL of 42 Degrees. I PROP. XXIV, and XXV. Theorems. F two Triangles (BAC, BAF) shall have two Sides (BA, AC) equal to two (BA, A F) each to each; and if one of the Triangles hath the Angle (B AF) contained by those Sides greater than the other (BAC); it shall have the Base BF greater than the Bafe (B C.) And again, If it hath the Base greater, it shall have the Angle greater. A From the Centre A describe a Circle which pafsfeth thro' C, it shall pass also thro' F, because A C, A Fare supposed to be equal. Therefore B F shall fall betwixt the Points A and C. Then join CF. The Angle B C F; is greater than the Angle ACF; that is, by the 5th of this Book, than the Angle A F C, and confequently much greater than the Angle BFC. There (a) Per19.1.1. fore in the Triangle BCF, (a) BF which is oppofite to the 2 the greater Angle BCF is greater than BC whch is oppofite to the leffer Angle B F C. 2. As for the Second Part of the Propofition this is manifest from the first Part. I PROP. XXVI. Theorem. F two Triangles (X and Z) have two Angles equal Fig. 25. to two, one Angle of the one equal to one Angle of the other (B to F and C to I), and one Side of one Equal to one of the other, whether it be that which is betwixt the equal Angles (as BC=FI) or a Side which is opposed to one of the equal Angles (as AC=LI); all the other Parts shall be equal. For first, let the Sides (BC, FI) which are betwixt the equal Angles be supposed equal: In this Cafe all the other Parts are equal; as hath been already demonftrated in the Scholium of Propofition 4. Again, suppose the Sides AC, LI which are opposed to the equal Angles to be equal. Here because the Angles (B, C) are by the Hypothesis equal to (F, I) the other Angles, also (A, L) shall be equal by Coroll. 9. Prop. 32. which Propofition depends not upon this. Therefore by the firit Part of this all the other Parts are equal. Coroll. Hence also, following Thales, we may mea- Fig. 84Sure inacceffible Distances. eg. Let AD be an inaccessible Line; to which at the Point A let there be erected the Perpendicular AC. Let there be made the Angle (ACB) equal to the Angle (ACD) the acceffible Line A B shall be equal to the inaccessible AD. 2. E. I. I PROP. XXVII, Theorem. F the right Line G O shall cut two right Lines which Fig. 45. are parallel (AB, CF); 1. The alternate Angles (RLO, QOL, likewise BLO, COL) shall be equal. 2. The external Angle GLB shall be equal to the internal one on the fame Side (that is, to LOF:) as likewise C3 GLR 1 Fig. 46. 12. GLRequal to LOC. 3. The two internal ones on the The first Part is thus proved. From O and L draw the Perpendiculars OR, LQ. These are perpendicu* Per. Ax.11. lar to the * two Parallels, A B, CF; and by Definition (a) Per Axio. 36, equal betwixt themselves, they shall therefore (a) intercept equal Parts of the Parallels, and R L shall be equal to QO. Therefore the Triangles X and Z are (b) Per 8.1.1. equilateral to one another. Therefore (b) the alternate Angles RLO, QOL which are oppofite to the equal Sides RO, QL are equal. Which is the first Thing. From whence it is likewise manifest that the Alternates BLO, COL are equal. For because as well BLO, (c) Per13.1.1. ALO as COL, FOL are equal (c) to two right ones: therefore BLO, ALO together, are equal to COL, FOL. Therefore taking away the Equals RLO, FOL, the remaining ones BLO, COL Thall be likewife equal. Part second. The Angle GLB is equal to that (d) Per15.1.1. which is vertically oppofite RLO (d); But RLO by the first Part of this Proposition is equal to LOF, Therefore GLB the external Angle is equal to the internal remote one which is on the fame Side, L O F. Part third. ALO by the first Part is equal to LOF. But LOF with COL make Angles equal to two right ones. Therefore A LO with COL doth the fame, Fig. 85. Coroll. Hence in Imitation of Eratosthenes we learn to measure the Compass of the Earth. For be observed that on the Day of the Summer Solstice, the Sun was perpendicularly over Siene, a City of Egypt; and be found by the means of a Stile perpendicularly erected, that on the Same Day the Sun was distant from the ver tical Point of Alexandria, a City of Egypt, fituate almost under the same Meridian with the other, seven Degrees, with one sth Part of a Degree; and he knew that these two Cities were about 5000 Furlongs distant from each other. From these Things by the Help of this Proposition be determin'd the Compass of the Earth. Let A be Siene, and B be Alexandria, where the Gnomon BC is erected perpendicular to the Horizon. Let DF and EG be the Solar Rays parallel to one another as to Sense. DA a Ray perpendicular to the Horizon of Siene; and EG a Ray Oblique to the Horizon of Alexandria, and which paffing by the Top of the Gnomon makes with it the Angle GCF, which is of 7 Degrees. Now seeing the Angle GCF is equal to the alternate one AFB, and the measure of it is the Arch AB of 7 Degrees; be found the Compass of the Earth by this Analogy; as 7 Degrees are to 5000 Furlongs; So the whole Circumference, which is of 360 Degrees, is in a gross Number to 250000, the Compass of the Earth in the same Measure. 2. E. I. PROP. XXVIII. Theorem. IF a right Line (GO) cutting two right Lines (AB, Fig. 47. CF) makes the alternate Angles (ALO, LOF) equal; the Lines (AB, FC) are parallel. If you deny it, let XLZ passing thro' the Point L be parallel to CF. Therefore XLO (a) is equal to (a) By the the alternate FOL, which cannot be, seeing by the foregoing. Hypothefis A LO is equal to FOL. PROP. XXIX. Theorem. IF a right Line GO cutting two right Lines (A B, Fig. 45, CF) shall make the external Angle (GL B) equal 46. to the internal oppofite one (LOF), or shall make the two internal Angles on the Same Side (ALO, COL) equal to two right Angles; (AB, CF) are parallel Lines, By the 15th of this Book GLB is equal to ALO, which is vertically oppofire to it. But by the Hypothefis GL B is equal to LOF. Therefore alfo ALO is equal to its alternate one LOF. Therefore (b) A B, (6) By the foregoing. CF'are parallel. Again, COL with FOL makes Angles equal to two right ones. But by the Hypothefis COL with ALO makes in all two right Angles also. Therefore ALO, FOL the alternate Angles are equal. Therefore, again, (c) A B, C F are parallel. C4 (c) By the Coroll, foregoing. 1 |