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CH A P. II.
The First Kudiments, or Leading and Preparatory

Problems, in Plain Geometry.

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Potulatcs or Petitions. 1. That a Right-line may be drawn from any one given Point to another.

2. That a Right line may be produced, encreased, or made longer from either of its Ends.

3. That upon any given Point (or Center) and with any given Distance (viz. with any Kadius) a Circle may be described.

PROBLEM I.
Two Right-lines being given, to find their Sum and

Diference. (3. 6. 1.)

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PROBLEM II.
To bifeet, or divide a Right-line given (as AB) into two

equal Parts (10. 2. 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 bileet the Line A B in the Middle at C; viz. it will make A G=CB; as was required.

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.. PROBLEM III. .. . To bifeet a Right-lin'd Angle given, into two equal Angles..

. . . . (9. e. 1.) : : . Upon thé Angular Point, as at C, with any convenient Radius, describe an Arch as AB; and from those Points A and B, describe two equal Arches crolling each other, as

at D; then join the Points C and · D with a Right-line, and it will

bifeet the Arch AB, and consequently the Angle; as was requir’d. '

PROBL E M IV. At a Point A, in a Right-line given AB, to make a Right-lin'd

Angle equal to a Right-lin'd Angle given C. (23. 6. 1.) Upon the given Angular Point C defcribe an Arch, as F D, (making C D any Radius at pleasure') and with the fame Radius describe the like Arch upon the given Point A, as fd; that is, make the Arch fd equal to the Arch FD; Then join the Points A and f with a Right-line, and it will form the Angle requir'd.'

PRO. PRO BLE, M v. To draw a Right-line, as FD, parallel to a given Right-line AB,

that shall pass thro' any align's Point, as at x, yiz. at any Distance requird. (31. 2. 1.)

Take any convenient Point in the given Line, as at C, (the farther off x the better;) make . . 11 .......***.......: C x Radius, and with it upon .F the Point C, describe a Semi-. circle, as HMxN; then make Athe Arch HM equal to the

AHCNB 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.

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

Right-Line, as at B; or upon any other Point align'd in it. (11.4, 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 C B 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 DB, and it will be the perpendicular as was Hequir'd.

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IRIT

· PROBLEM VIII. Ta divide any given Right-line, as A B, into any proposed 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 Points A and B, set off the proposed Number of equal Parts (Juppose '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 AB into the Number of equal Parts required.

PROBLEM IX. To describe a Circle that all 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 bileet both those Lines (per Problem 2.) the Point where the bisecting Lines meet, as at C, will be the Center of the Circle was required.

The Work of this Problem being well understood, 'twill be easy to perform the two following, without any Scheme, viz.

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1. To find the Center of any Circle given. (1. 4. 3.) By the last Problem ’is plain, that if three Points be any where taken in the given Circle's Periphery, as at A, B, D, the Center of that Circle may 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 Peints in the given Sege ment's Arch, and then proceed as before.

PRO

PROBLEM X. · Upon a Right-line given, as A B, to describe an Equilateral

Triangle. (1. 2. I.) 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. A C and B C; then join the Points A and B C with Rightlines, and they will make the Triangle requir'd.

i. PROBLEM XI. 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.)

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Make either of the porter Lines (as AC) Radius, and upon either End of the longest Line (as at A) describe an Arch; then make the other Line C B Radius, and upon the a ther End of the longest Side (as at B) describe another Arch, to cross the First Arch (as.at C): Join the Points C A and C B with Right-lines, and they will form the Triangle required.

. PROBLEM XII. Upon a given Right-line, as A B, to form a Square. (46.6. 1.)

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

PRO

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