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

: angled Parallelogram.

Let the given Lines be

BC Upon one End of the longest Line, as at B, erect a Perpendicular of the same Length with the fortest Line BC; then from the Point C draw a Line parallel, 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 Paral. lelograms, they are made, or described, after the same Manner with the two laft Figures ; only instead of erecting the Perpendiculars, you must set off their given Angles, and then proceed to draw their Sides parallel, &c. as before.

PROBLEM XIV. in any given Circle, to inscribe or make a Triangle, whose Angles salt

be equal to the Angles of a given Triangle ; as the Triangle FDG. (2. e. 4.)

Note, Ang Right-lin'd Figure is said to be inscrib'd in a Circle, when all the Angular Points of that Figure do juft touch the Circle's Periphery.

Draw any Right-line (as HK) so as just to touch the Circle, as at A; then make the Angle

НА K AC equal to any one Angle of the given Triangle, as DFG; and the Angle H AB equal to another Angle of the Triangle, as DG F; then will the Angle BAC be equal to the Angle FDG. Join the Points B and C with a Right-line, and 'twill form the Triangle required. .

PRO

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298

Elements of Geometry.
Elements of 6

Part III.

PROBLEM XV. In any given Triangle, as A BD, to describe a Circle that fball

touch all its Sides. (4. 6. 4.) Bisedt any two Angles of the Triangle, as A and B, and where the bisecting 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. To describe a Circle about any given Triangle. (5. e. 4.) This Problem is perform'd in all respects like the Ninth, viz. by bisecting any Two Sides of the given Triangle ; the Point, where those bisecting Lines meet, will be the Center of the Circle required.

PROBLEM XVII.
To describe a Square about any given Circle, (7.1.4.)
Draw two Diameters in the given i f
Circle (as D A and EB) crossing ati
Right Angles in the Center C; and,
with the Circle's Radius C A, describe
from the extream Points of those Dia. D
ineters, viz. A, B, D, E, cross 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.

PROBLÉ M XVIII.
In any given Circle, to describe the largest Square it can

contain. (6. l. 4.) Having drawn the Diameters, as D A and EB, bisecting each other at Righe-angles in the Center C, (as in the last Scheme); then join the Points A, B, D, and Ē, with Right-lines, viz. A B, B D, DE, E A, and they will be Sides of the Square ses quired.

: PRO

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PROBLEM XX.
In any given Circle, to describe a regular Pentagon.

(11.2. 4. & 10. 2. 3.). Or, in general Terms, to describe any regular Polygon in a

Circle. Draw the Circle's Diameter DA, and divide it into as many equal Parts as the proposed Polygon hath Sides ; then make the whole Diameter a Radius, and describe the two Arches CA and CD. If a Right-line be drawn from the Point C, through the Second of those equal Parts in the Diameter; as at 2, it will assign, a Point in the oppolite Semicircle's Periphery, as at B. Join DB with a Right-line, and it will be the Side of the Pentagon required.

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These Twenty Problems are sufficient to exercise the young Practitioner, and bring his Hand to the right Management of a Ruler and Compasses, wherein I would advise him to be very ready and exact. .

As to the Reason why such Lines must be so drawn as directed at each 'Problem, that, I presume, will' fully and clearly appear from the following Theorems ; and therefore I have (for Brevity's Sake) omitted giving any Demonstrations of them in this Chapter, defiring the Learner to be satisfied with the bare Knowledge of doing them only, until he hath fully considered the Contents of the next Chapter; and then I doubt not but all will appear very plain and easy.

СНА Р. ІІІ. A Colleation of most useful Theorems in plain Geometry

. Demontraist. Note, In order to shorten several of the following Demonftrationsa

it will be necessary to premise, that 1. THE Periphery (or Circumference) of every Circle (whe

ther great or small) is suppos’d to be divided into 360 equal Parts, called Degrees; and every one of those Degrees are divided into 60 equal Parts, callid Minutes, &c.

2. All Angles are measured by the Arch of a Circle describ'd upon the Angular Point (See Defin. 9. Page 287.) and are esteemid greater or less, 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 those five general Axioms already laid down in Page 146, (which I here suppose 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.

nent de ery real

de

Arioms.
6. Every whole Thing is Greater than its Part.
That is, the whole Line A B is
greater than its Part A c, &c. Ama
The same is to be understood of Superficies's and Solids.

7. Every Whole is equal to all its Parts taken together.
That is, the whole Line A B is equal é d e
to its Parts AC +cd+dete B. SA-1-1-1-B

The same is also true in Super ficies's and Solids.

8. Those Things which, being laid one upon another, do agret or meet in all their Parts, are equal one to the other.

But the Converse of this Axiom, to wit, that equal Things be. ing 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 same Figure or Form: As for Instance, 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. Also, 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, Besides the Characters already explain'd in Part I, and in

other Places of this Tract, these following are added. Viz. 5 denotes an Angle in general, and a signifies Angles; a signifies a Triangle; o signifies a Square, and denotes a Parallelogram. And when an Angle is denoted by any three Letters (as, A B C) the middle Letter (as B) always denotes the Angular Point; and the other two Letters (as A E, and B C) denote the Lines or Sides of the Triangle which includes that Angle.

These Things being premised, the young Geometer may proceed to the Demonstrations of the following Theorems ; wherein he may perceive an absolute Neceflity of being well versed in several Things that have been already deliver’d: And also it will be very advantageous to store up several useful Corollaries and Lemma's, as they become discover'd Truths: For it often happens, that a Proposition cannot be clearly demonstrated a priori, or of itself, without a great Deal of Trouble; therefore it will be useful to have Recourse to those Truths that may be affifting in the Demonstration then in Hand.

THE O..

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