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PROBL E M' II. To bifect, or divide a Right-line given. (as AB) into two

equal Parts (10. 6. 1.) From both Ends of the given Line (viz. A and B) with any - Radius greater than half its Length, describe Two Arches that may_cross each other in two Points, as at D and F; then join those Points D F with a Right-line, and it will bifect the Line AB in the Middle at C; viz. it will make AC=CB; as was required.

PROBL E M III. To bifezt a Right-lin'd Angle given, into two equal Angles,

(9: 6. 1.) Upon the Angular Point, as at C, with any convenient Radius, describe an Arch as A B; and from

А those Points A and B, describe tivo equal Arches crofing each other, as at D; then join the Points C and

D D with a Right-line, and it will biselt the Arch AB, and consequent

B ly the Angle; as was requir’d.

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PROBLEM IV. At a Point A, in a Right-line given A B, to make a Right-lin'd

Angle equal to a Right-lin'd Angle given C. (23. 1. 1.) Upon the given Angular Point C describe an Arch, as F D, (making CD any Radius at pleasure) and with the fame Radius describe the like Arch upon the given Point A, as fd; that is,

D make the Arch f d equal to the Arch FD; Then join the Points A and with a Right-line, and it will form the Angle requir’d.

A .



PROBLEM V. To draw a Right-line, as FD, parallel to a given Right-line AB,

that Mall pass thro' any affign'd Point, as at x, viz. at any Distance requir'd. (31.1. 1.),

Take any convenient Point in the given Line, as at C; (the farther off * the better;) make

M. Cx Radius, and with it upori F

D the Point C, describe a Semicircle, as HMxN; then make A

e the Arch HM equal to the


N Arch * N; thro' the Points M and x draw the Right-line FD, and it will be parallel to the Line AC, as was requir'd.


To let fall a Perpendicular, as C*, upon a given Right-line AB, from any angn’d Point that is not in it, as from C. (12. 2. 1:)

Upon the given Point C describe such an Arch of a Circle as will cross the given Line A B in two

с Points; as at d and f; Then biseet the Distance between those two Points df (per Probl. 2.) as at x. Draw the Right-line C x, and it will be the Perpendicular requir’d.


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To erect or raise a Perpendicular upon the End of any given

Right-Line, as at B; or upon any other Point asign' in it. (11.6. 1.) Upon any Point (taken at an Adventure) out of the given Line, as at C, describe such a Circle as will pass through the point from whence the perpendicular must be raised, as at B, (viz. make CB Radius): And from the point where the Circle cuts the given Line, as at A, draw the Circle's Diameter AC D; then from the Point D draw the Right-line D B, and it will be the Perpendicular as was requir'd.



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To divide any given Right-line, as A B, into any propofed Number of

equal Parts. (10. 6. 6.) At the extream Points (or Ends) of the given Line, as at A and B, make two equal Angles (by Prob. 4.) continuing. their Sides A D and B C to any sufficient Length; then upon those Sides, beginning at the

B Points A and B, set off the proposed Number of equal Parts

I (suppose 'em 5.) If Right-lines be drawn (cross the given Line) from one point to the other, as in the annexed Figure, those Lines will divide the given Line A B into the Number of equal Parts required.

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To describe a Circle that shall pass (or cut) thro' any Three
Points given, not lying in a Right-line, as at the Points

A B D.
Join the Points A B and B D with Right-lines; then bisect
both those Lines (per Problem 2.) the

Point where the bisecting Lines meet,
ay at C, will be the Center of the Circle
The Work of this problem being well

understood, 'twill be easy to perform the
two following, without any Scheme, viz.



1. To find the Center of any Circle given. (1. 1. 3.) By the last Problem 'tis plain, that if three Points be any where taken in the given Circle's Periphery, as at A, B, D, the Center of chat Circle

be found as before.


2. If a Segment of any Circle be given, to compleat or describe

the whole Circle. This may be done by taking any three Paints in the given Sege pent's Arch, and then proceed as before.


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Upon a Right-line given, as A B, to describe an Equilateral

Triangle. (1. l. 1.)
Make the given Line Radius, and
with it, upon each of its extream Points
or Ends, as at A and B, describe an
Arch, viz. AC and B C; then join
the Points A and B C with Right-
lines, and they will make the Triangle


Three Right-lines being given, to form them into a Triangle,

(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 4) describe an
Arch; then make the other Line CB Radius, and upon the o-
ther End of the longest Side (as at B) describe another Arch, to
cross the First Arch (as at C): Join the Points CA and C B with
Right-lines, and they will form the Triangle required.

P R 0 B L E M XII.
Upon a given Right-line, as AB, to form a Square. (46. e. 1.)

Upon one End of the given Line, as at B, erect the Perpendicular B D, equal in Length with the

D given Line, viz. make BD=AB; that being done, make the given Line Radius, and upon the Points A and D describe equal Arches to cross each other, as at C; then join the Points C A and CD with Right-lines, and they will


3 form the Square required.



Teva unequal Right-lines being given, form or make of thein a Right-

angled Parallelogram.

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Let the given Lines be

Upon one End of the longest

Line, as at B, erect a Perpendi-
cular of the fame Length with
the shortest Line BC; then from


B the Point G draw a Line parallel, and of the same 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.

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In any given Circle, to inscribe or make a Triangle, whose Angles Mall

be equal to the Angles of a given Triangle ; as the Triangle FDG. (2. 6. 4.)

Note, Ang Right-lind Figure is said to be infcrib'd in a Circle, when all the Angular Points of that Figure do just 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


KAC equal to any one Angie

of the given Triangle, as DFG;
and the Angle H A B equal to
another Angle of the Triangle,
as DGF; then will the Angle
BAC be equal to the Angle
FDG. Join the Points B and
C with a Right-line, and 't will
form the Triangle required,

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