PROBLEM. IL

To bifect, or divide a Right-line given (as AB) into twe equal Parts (10. e. 1.)

.D.

From both Ends of the given Line (viz. A and B) with any Radius greater than half its Length, defcribe Two Arches that may cross each other in two Points, as at D and F; then join thofe Points D F with à Right-line, and it will bifect the Line AB in the Middle at C; viz. it will make AC=CB; as was required.

A

PROBLEM III.

To bifect a Right-lin❜d Angle given, into two equal Angles,

(9. e. 1.)

Upon the Angular Point, as at C, with any convenient Radius,

defcribe an Arch as AB; and from those Points A and B, defcribe two equal Arches croffing each other, as at D; then join the Points C and D with a Right-line, and it will bifelt the Arch AB, and confequently the Angle; as was requir'd

PROBLEM IV.

A

D

B

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. e. 1.)

Upon the given Angular Point C deferibe an Arch, as FD, (making CD any Radius at pleasure) and with the fame Radius defcribe 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 ƒ with a Right-line, and it will form the 'Angle requir❜d.

D

PRO

PROBLEM V.

To draw a Right-line, as FD, parallel to a given Right-line AB, that shall pass thro' any affign'd Point, as at x, viz. at any DiStance requir'd. (31. e. 1.),

F

M.........

·D

N

B

Take any convenient Point in the given Line, as at C, (the farther off x the better;) make Cx Radius, and with it upon the Point C, defcribe a Semicircle, as HMx N; then make the Arch HM equal to the Arch x N; thro' the Points M and x draw the Right-line FD, and it will be parallel to the Line A C, as was requir’d.

A

H

PROBLEM VI.

To let fall a Perpendicular, as Cx, upon a given Right-line AB, from any affign'd Point that is not in it, as from C. (12. e. 1;) Upon the given Point C deferibe fuch an Arch of a Circle as will cross the given Line AB in two Points, as at d and f; Then bifect the Distance between thofe two Points df (per Probl. 2.) as at x. Draw the Right-line Cx, and it will be the Perpendicular requir'd.

A

PROBLEM VII.

L

To erect or raife a Perpendicular upon the End of any given Right-Line, as at B; or upon any other Point affign'd in it. (11. e. 1.)

C

Upon any Point (taken at an Adventure) out of the given Line, as at C, defcribe fuch a Circle as will pass through the Point from whence the Perpendicular must be raifed, 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 D B, and it will be the Perpendicular as was

requir❜d.

PRO

PROBLEM VIII.

To divide any given Right-line, as A B, into any propofed Number of equal Parts. (10. e. 6.)

2

I

B

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 AD and B C to any fufficient Length; then upon thofe Sides, beginning at the Points A and B, fet off the propofed Number of equal Parts (Suppose 'em 5.) If Right-lines be drawn (cross the given Line) from one Point to the other, as

2

3

་་་་་་་

D

in the annexed Figure, thofe Lines will divide the given Line A B into the Number of equal Parts required.

PROBLEM IX.

To defcribe 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.

A

Join the Points A B and B D with Right-lines; then bisect both thofe Lines (per Problem 2.) the Point where the bifecting Lines meet, aş at C, will be the Center of the Circle required.

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

B

D

to

1. To find the Center of any Circle given. (1. e. 3.)

By the laft 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 that Circle may be found as before.

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

the whole Circle.

This may be done by taking any three Paints in the given Seg

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. e. 1.)

Make the given Line Radius, and with it, upon each of its extream Points or Ends, as at A and B, defcribe an Arch, viz. A C and B C; then join the Points AC and B C with Rightlines, and they will make the Triangle requir❜d.

A

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.)

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 cross the First Arch (as at C): Join the Points CA and C B with Right-lines, and they will form the Triangle required.

PROBLEM XII.

Úpon a given Right-line, as AB, to form a Square. (46. e. 1.)

D

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 crofs each other, as at C; then join the Points CA and CD with Right-lines, and they will form the Square required.

A

PRO

PROBLEM XIII.

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

lel, and of the fame Length, to AB; viz. make DC= AB: Join DA 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 set off their given Angles, and then proceed to draw their Sides parallel, &c. as before.

PROBLEM XIV.

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. e. 4.)

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

D

H

G

A

K

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 Angle of the given Triangle, as DFG; and the Angle HAB 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