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H, and (by prob. 8.) draw HK parallel thereto, so will Ell be the fourth proportional required.

For, by cor. 1. theo. 20. EG : EI::EH: EK.

Or, A:B::C: EK.


Pl. 3. fig. 1.

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

Draw an indefinite blank line, as AF, on which lay the line A, from A to B, and the line B, from B to C, on the point B, which is the joining point of the lines A and B ; erect a perpendicular BD (by prob. 5.) bisect AC in E (by prob. 4.) and describe the semicircle ADC; and from the point D, where the periphery cuts the perpendicular BD, draw the line BD, and that will be the mean proportional required.

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 BBC, 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.


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 iwo 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 cor. 1. theo. 20. AC: AD:: AE: AB. Or, C:D :: AE: EB.


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.


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


PL. 3. fig. 4.

To make an angle of any number of degrees, at the point n, 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 À 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.


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 ed, measured on the same line of chords, gives the quantity of the angle BAC, as required; this is plain from def. 17.


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 , and ECB will be the triangle required. For the triangles ADC, AEC, being upon

the same base AC, and under the same parallel ED, (by cor. to theo. 13.) will be equal, therefore if ABC be added to each, then ABCD =BEC.


PL. 3. fig. 7,

To make a triangle DFH, equal to a given five-sided figure ABCDE.

Draw DA and DB, and also EH and CF, parallel to them (by prob. 8.) meeting AB produced in H and F; then draw DH, DF, and the triangle HDF is the one required.

For the triangle DEA= DHA, and DBC = DFB (by cor. to theo. 13.) therefore by adding these equations, DEA + DBC= DHĂ+ DFB if to each of these ADB be added; then DEA+ ADB + DBC= ABCDE = (DHA+ ABD+ DFB, = DHF,



PL. 3. fig. 8.

To project the lines of chords, sines, tangents and sccante, with any radius.


On the line AB, let a semicircle ADB be de scribed ; let CDF be drawn perpendicular to this line from the centre C; and the tangent BE perpendicular to the end of the diameter; let the quadrants, AD, DB, be each divided into 9 equal parts, every one of which will be 10 degrees; if then from the centre C, lines be drawn through 10, 20, 30, 40, &c. the divisions of the quadrant BD, and continued to BE, we shall there have the tangents of 10, 20, 30, 40, &c. and the secants C 10, Č 20, C30, &c. are transferred to the line CF, by describing the arcs 10, 10: 20, 20: 30, 30, &c. If from 10,

20, 30, &c. the divisions of the quadrant BD, there - be let fall perpendiculars, let these be transferred to the radius CB, and we shall have the sines of 10, 20, 30, &c. and if from A we describe the arcs 10, 10: 20, 20:30, 30, &c. from every division of the arc AD; we shall have a line of chords. The same way we may have the sine, tangent, &c. to every single degree on the quadrant, by şubdividing each of the 9 former divisions into 10 equal parts. By this method the sines, tangents, & c. may be drawn to any radius ; and then, after they are transferred to lines on a rule, we shall have the scales of sines, tangents, &c. ready for use.



The strictness of geometrical demonstration admits of no other instruments, than a rule and a pair of compasses. But, in proportion as the practice of geometry was extended to the different arts, either connected with, or dependent upon it, new instruments became necessary, some to answer peculiar

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