PROBLEM XV. In any given Triangle, as ABD, to defcribe a Circle that fall touch all its Sides. (4. e. 4.) Bifect any two Angles of the Triangle, as A and B, and where the bifecting Lines meet (as at C) will be the Center of the Circle required; and its Radius will be the nearest Distance to the Sides of the Triangle. PROBLEM XVI. B To defcribe a Circle about any given Triangle. (5. e. 4.) D This Problem is perform'd in all refpects like the Ninth, viz. by bifecting any Two Sides of the given Triangle; the Point, where thofe bifecting Lines meet, will be the Center of the Circle required. PROBLEM XVII. F E G To defcribe a Square about any given Circle. (7. e. 4.) Draw two Diameters in the given Circle (as D A and E B) croffing at Right Angles in the Center C; and, with the Circle's Radius C A, defcribe from the extream Points of thofe Diameters, viz. A, B, D, E, crofs Arches, as at F, G, H, K; then join those Points where the Arches cross with Right Lines, and they will form the Square required. PROBLEM XVIII. In any given Circle, to defcribe the largest Square it can A Having drawn the Diameters, as D A and E B, bifecting each other at Right-angles in the Center C, (as in the laft Scheme); then join the Points A, B, D, and E, with Right lines, viz. A B, B D, DE, E A, and they will be Sides of the Square re quired, PRO PROBLEM XIX. Upon any given Right-line, as AB, to defcribe a regular Pentagon, or Five-fided Polygon. Make the given Line Radius, and upon each End of it de fcribe a Circle; and through those Points where the Circles cross each other (as at G x) draw the Rightline Gex: Upon the Point G with the fame Radius defcribe the Arch HAB D, and laying a Ruler upon the Points D, e, mark where it croffes the other Circle, as at F. Again, lay the Ruler upon the Points H, e, and mark where it croffes the other Circle, as at C: Then from H x C B ....... the Points F and C (with the fame Radius as before) defcribe crofs Arches, as at K: Join the Points AF, FK, KC, and C B, with Right-lines, and they will form the Pentagon required, viz. AF FKKC CB AB; and the Angles at A, B, C, K, F will be equal. PROBLEM XX. In any given Circle, to defcribe a regular Pentagon. Or, in general Terms, to defcribe any regular Polygon in a Circle. C... Draw the Circle's Diameter D A, and divide it into as many equal Parts as the propofed Polygon hath Sides; then make the whole Diameter a Radius, and defcribe the two Arches CA and CD. If a Right-line be drawn from the Point C, through the Second of thofe equal Parts in the Diameter, as at 2, it will affign a Point in the oppofite Semicircle's Periphery, as at B. Join DB with a Right-line, and it will be the Side of the Pentagon required. Q92 D A 2 3 4 Thefe Thefe Twenty Problems are fufficient to exercise the young Practitioner, and bring his Hand to the right Management of a Ruler and Compaffes, wherein I would advise him to be very ready and exact. As to the Reason why fuch Lines must be fo drawn as directed at each Problem, that, I prefume, will fully and clearly appear from the following Theorems; and therefore I have (for Brevity's Sake) omitted giving any Demonftrations of them in this Chapter, defiring the Learner to be fatisfied with the bare Knowledge of doing them only, until he hath fully confidered the Contents of the next Chapter; and then I doubt not but all will appear very plain and easy. CHA P. III. A Collection of most useful Theorems in plain Geometry Demonftrated, Note, In order to shorten feveral of the following Demonflrations, it will be neceffary to premife, that 1.THE Periphery (or Circumference) of every Circle (whether great or small) is fuppos'd to be divided into 360 equal Parts, called Degrees; and every one of thofe Degrees are divided into 60 equal Parts, call'd Minutes, &c. 2. All Angles are measured by the Arch of a Circle defcrib'd upon the Angular Point (See Defin. 9. Page 287.) and are esteem'd greater or lefs, according to the Number of Degrees contain❜d in that Arch. 3. A Quadrant, or Quarter part of any Circle, is always go Degrees, being the Measure of a Right-angle (Defin. 6. P. 287.) and a Semicircle is 180 Degrees, being the Measure of two Right-angles. 4. Equal Arches of a Circle, or of equal Circles, measure equal Angles. To thofe five general Axioms already laid down in Page 146, (which I here fuppofe the Reader to be very well acquainted with) it will be convenient to understand these following, which begin their Number where the other ended. Arioms. Axioms. 6. Every whole Thing is Greater than its Part. That is, the whole Line AB is greater than its Part A c, &c. is } ^ A The fame is to be understood of Superficies's and Solids. C d B 7 Every Whole is Equal to all its Parts taken together. That is, the whole Line AB is equal to its Parts AC+cd+de+e B. $ A—|—|—|—B The fame is alfo true in Superficies's and Solids. -B 8. Thofe Things which, being laid one upon another, do agree. or meet in all their Parts, are equal one to the other. But the Converfe of this Axiom, to wit, that equal Things being laid one upon the other will meet, is only true in Lines and Angles, but not in Superficies's, unless they be alike, viz. of the fame Figure or Form: As for Inftance, a Circle may be equal in Area to a Square; but if they are laid one upon the other, 'tis plain they cannot meet in all their Parts, because they are unlike Figures. Alfo, a Parallelogram and a Triangle may be equal in their Area's one to another, and both of them may be equal to a Square; but if they are laid one upon the other, they will not meet in all their Parts, &c. Note, Befides the Characters already explain'd in Part I, and in other Places of this Tract, thefe following are added. Viz. denotes an Angle in general, and fignifies Angles; A fignifies a Triangle; fignifies a Square, and denotes a Parallelogram. And when an Angle is denoted by any three Letters (as, ABC) the middle Letter (as B) always denotes the Angular Point; and the other two Letters (as A B, and B C) denote the Lines or Sides of the Triangle which includes that Angle. Thefe Things being premifed, the young Geometer may proceed to the Demonftrations of the following Theorems; wherein he may perceive an abfolute Neceffity of being well versed in feveral Things that have been already deliver'd: And alfo it will be very advantageous to ftore up feveral ufeful Corollaries and Lemma's, as they become difcover'd Truths: For it often happens, that a Propofition cannot be clearly demonftrated a priori, or of itfelf, without a great Deal of Trouble; therefore it will be ufeful to have Recourfe to thofe Truths that may be affifting in the Demonftration then in Hand. THE O THEOREM I. If a Right line ftand upon (or meet with) another Right-line, and make Angles with it, they will either be two Right-angles, or two Angles equal to two Right-angles. (13. e. 1.) Demonftration. Suppose the Lines to be AB and D C, meeting in the Point at C: Upon C defcribe any Circle at pleasure; then will the Arch A D be the Measure of the b, and the Arch D B the Measure of but the Arches A D+D B=180°, viz. they compleat the Semicircle. be C Confequently the <b+<= 180°. Which was to be prov'd. Corollaries. 1. Hence it follows, that if the b90° thene 90°; but ifb be obtufe, then thee will be acute, &c. From hence it will be eafy to conceive, that if feveral Rightlines ftand upon, or meet with any Right-line at one and the fame Point, and on the fame Side, then all the Angles taken together will be 180°, viz. Two Right-angles. THEOREM II. If two Angles interfect (i. e. cut or cross) each other, the two op¬ pofite Angles will be equal. (15. e. 1.) Demonftration. Let the two Lines be A B and DE, interfecting each other in the Then be=180°2 Subtract on both Sides of the Equation, and it will leave <e=<a. b/e a/c Again, be=180°, as before; and <+<c= 180°, confequently <e+C=<b<e. Subtract <e, and then C6. Q. E. D. Carol |