PROBLEM X.

Upon a Right-line given, as AB, to defcribe an Equilateral Triangle. (1. e. 1.)

Make the given Line Radius, and with it, upon each of its extream Points or Ends, as at A and B, defcribe the Arches AC and BC; join their Interfection C with the Points A and B, then will ACB be the Triangle required.

B

PROBLEM XI.

To form a Triangle of three given Right-lines, (provided any two of them, taken together, be longer than the Third.) (22. e. 1.)

Let the given Lines be

Make either of the shorter Lines (as AC) Radius, and upon either End of the longest Line (as at A) defcribe an

Arch; then make the other Line CB Radius, and upon the other End of the longeft Side (as at B) defcribe another Arch, to crofs the First Arch (as at C): Join the Points CA and CB with Right-lines, and they will form the Triangle required.

PROBLEM XII.

Upon a given Right-line, as A B, to defcribe a Square. (46. e. 1.)

D

Upon one End of the given Line, as at B, erect the Perpendicular BD, equal in Length with the given Line, viz. make BD=AB; that being done, make the given Line Radius, with which upon the Points A and D defcribe Arches to crofs each other, as at C; then join the Points C, A and alfo C, D with Right-lines, and they will form the Square required.

A

PROBLEM XIII.

Two unequal Right-lines being given, to form or make of them a Right-angled Parallelogram.

Let the given Lines be

Upon the End of the longest Line, as at B, erect a Perpendicular of the fame Length with the fhorteft Line BC; then from -the Point C draw a Line paral

lel, and of the fame Length, to AB; viz. make DC AB: Join D A with a Right-line, and it will form the Oblong or Parallelogram required.

As for Rhombus's and Rhomboides's, to wit, Oblique-angled Parallelograms, they are made, or defcribed, after the fame Manner with the two laft Figures; only inftead of erecting the Perpendiculars, you must fet off their given Angles, and then proceed to draw their Sides parallel, &c. as before.

In

any given Circle, to infcribe or make a Triangle, whofe Angles fhall be equal to the Angles of a given Triangle; as the Triangle FDG. (2. c. 4.)

Note, Any Right-lined Figure is faid to be infcrib'd in a Circle, when all the Angular Points of that Figure do just touch the Circle's Periphery.

D

k

Draw any Right-line (as HK) fo as juft to touch the Circle, as at A; then make the Angle KAC equal to any one Angle of the given Triangle, as DFG; and the Angle HAB equal to another Angle of the Triangle, as DGF; then will the Angle BAC be equal to the Angle FDG. Join the Points Band Cwith a Right-line, and 'twill compleat the Triangle required.

Qq

703

PRO

In any given Triangle, as ABD, to defcribe a Circle that ball 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 'requi red; and its Radius will be the Perpendicular let fall from C, upon either Side 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 those bilecting Lines meet, will be the Center of the Circle required.

PROBLEM XVII.

To defcribe a Square about any given Circle. (7. e. 4.)

Draw two Diameters in the given Circle (as DA 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 those 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.

F

E

G

D

A

C

HA

PROBLEM XVIII.

In any given Circle, to defcribe a Square. (6. e. 4.) Having drawn the Diameters, as D A and E B, bifecting each other at Right-angies in the Center C, (as in the last Scheme); then join the Points A, B, D, and E, with Right lines, viz. AB, BD, DE, E A, and they will be Sides of the Square, as required. PR Q

PROBLEM XIX.

Upon any given Right-line, as AB, to defcribe a regular Pentagon, or Five-fided Polygon.

K.

F

B

Make the given Line Radius, and upon each End of it defcribe 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 HA BD, 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 crosses the other Circle, as at C: Then from the Points F and C (with the fame Radius as before) defcribe cross 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 FK KC=CB= AB; and the Angles at A, B, C, K, F will be equal.

=

H

PROBLEM XX.

In any given Circle, to describe a regular Pentagon.
(11. e. 4. & 10. e. 3.)

Or, in general Terms, to defcribe any regular Polygon in a
Circle.

Draw the Circle's Diameter D A, and divide it into as many

equal Parts as the proposed 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

C

I

2 3 4

B

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 Reafon why fuch Lines muft be fo drawn as directed at each Problem, that, I preíume, 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 Theozems in plane Geometry Demonftrated.

I.

Note, In order to shorten feveral of the following Demonflrations, Ait will be necessary to premife, that

ΤΗ HE Periphery (or Circumference) of every Circle (whether great or fmall) 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 90 Degrees, being the Meajure of a Right-angle (Defin, 6. P. 287.) and a Semicircle is 180 Degrees, being the Measure of twa Right angles.

4. Equal Arches of a Circle, or of equal Circles, meafure 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 underftand thefe following, which begin their Number where the other ended.

Arioms.