Make BC perpendicular and equal to AB; and from A and C, with the line AB, or BC, let two arcs be described, cutting each other in D; from whence to A and C, let the lines AD, DC be drawn; so is ABCD the square required.

For all the sides are equal by construction; therefore the triangles ADC and BAC, are mutually equilateral and equiangular, and ABCD is an equilateral parallelogram, whose angles are right. For B being right, D is also right, and DAC, DCA, BAČ, AČB, each half a right angle, (by lemma preceding theo. 7. and cor. 2. theo. 5.) whence DAB and BCD will each be a right angle, and (by def. 44.) ABCD is a

square.

SCHOLIUM.

By the same method a rectangle or oblong, may be described, the sides thereof being given.

PROB. X.

PL. 2. fig. 15.

To divide a given right line AB into any proposed number of equal parts.

Draw the indefinite right line AP, making any angle with AB, also draw BQ parallel to AP, in each of which, let there be taken as many equal parts AM, MN, &c. Bo, on, &c. as you would have AB divided into; then draw Mm, Nn, &c. intersecting AB in E, F, &c. and it is done.

For MN and mn being equal and parallel, FN will be parallel to EM; and in the same manner, GŌ to FN (by theo. 12.) therefore AM,

MN, NO, being all equal by construction, it is plain (from theo. 10.) that AE, EF, FG, &c. will likewise be equal.

PROB. XI.

PL. 2. fig. 16.

To find a third proportional to two given right lines, A and B.

Draw two indefinite blank lines CE, CD, anywise to make any angle. Lay the line A, from C to F; and the line B, from C to G; and draw the line FG; lay again the line A, from C to H; and through H draw HI parallel to FG (by prob. 8.) so is ČI the third proportional required.

For by cor. 1. theo. 20, CG: CH:: CF: CI. Or, B: A:: A: CI

PROB. XII.

PL. 2. fig. 17.

Three right lines A, B, C, given to find a fourth proportional.

Having made an angle DEF anywise, by two indefinite blank right lines ED, EF, as before; lay the line A, from E to G; the line B, from E to I; and draw the line IG; lay the line C, from E to H, and (by prob. 8.) draw HK parallel thereto, so will EK be the fourth proportional required.

For, by cor. 1. theo. 20. EG: E1: : EH ; EK. Or, A B C : EK.

PROB. XIII.

PL. 3. fig. 1.

Two right lines, A and B, given to find a mean proportional.

Draw an indefinite straight line, on which place ABA, and BC= B; bisect AC (by prob. 4.) in E, and describe the semicircle ADC; and from the point B, erect the perpendicular BD, (by prob. 5.) then BD is a mean proportional.

For if the lines AD, DC, be drawn, the angle ADC is a right angle (by cor. 5. theo. 7.) being an angle in a semicircle.

The angles ABD, DBC, are right ones (by def. 10.) the line BD being a perpendicular; wherefore the triangles ABD, DBC, are similar: thus the angle ABD = DBC, being both right, the angle DAC is the complement of BDA to a right angle (by cor. 2. theo. 5.) and is therefore equal to BDC, the angle ADC being a right angle as before; consequently (by cor. 1. theo. 5.) the angle ADB = DCB, wherefore (by theo. 16.)

AB : BD : : BD ; BC.

Or, A : BD : : BD : B,

PROB. XIV.

PL 3. fig. 2.

To divide a right line AB, in the point E, so that AE shall have the same proportion to EB, as two given lines C and D have.

Draw an indefinite blank line, AF, to the extremity of the line AB, to make with it any angle; lay the line C, from A to C; and D, from C to D; and join the points B and D, by the line

BD; through C draw CE parallel to BD (by prob. 8.) so is E the point of division.

For, by theo. 20. AC: CD::AE: EB.

Or, C: D:: AE ; EB.

PROB. XV.

PL. 3. fig. 3.

To describe a circle about a triangle ABC, or (which is the same thing) through any three points, A, B, C, which are not situated in a right line.

By prob. 4. Bisect the line AC by the perpendicular DE, and also CB, by the perpendicular FG, the point of intersection H, of these perpendiculars, is the centre of the circle required; from which take the distance to any of the three points A, B, C, and describe the circle ABC, and it is done.

For, by cor, to theo. 8. The lines DE and FG, must each pass through the centre, therefore, their point of intersection H, must be the

centre.

SCHOLIUM.

By this method the centre of a circle may be found, by having only a segment of it given.

PROB. XVI.

PL. 3. fig. 4.

To make an angle of any number of degrees, at the point A, of the line AB, suppose of 45 degrees.

From a scale of chords take 60 degrees, for 60° is equal to the radius, (by cor. theo. 15.) and with that distance from A, as a centre, describe a circle from the line AB; take 45 degrees, the

quantity of the given angle, from the same scale of chords, and lay it on that circle from a to b; through A and b, draw the line AbC, and the angle A will be an angle of 45 degrees, as required.

If the given angle be more than 90°, take its half (or divide it into any two parts less than 90) and lay them after each other on the arc, which is described with the chord of 60 degrees; through the extremity of which, and the centre, let a line be drawn, and that will form the angle required, with the given line.

PROB. XVII.

PL. 3. fig. 5.

To measure a given angle, ABC.

If the lines which include the angle, be not as long as the chord of 60° on your scale, produce them to that or a greater length, and between them so produced, with the chord of 60°, from B, describe the arc ed; which distance e d, measured on the same line of chords, gives the quantity of the angle BAC, as required; this is plain from def. 17.

PROB. XVIII.

PL. 3. fig. 6.

To make a triangle BCE equal to a given quadrilateral figure ABCD.

Draw the diagonal AC, and parallel to it (by prob. 8.) DE, meeting AB produced in E; then draw CÉ, and ECB will be the triangle required.

For the triangles ADC, AEC, being upon the same base AC, and under the same parallel ED,