GEOMETRICAL PROBLEMS.

PROBLEM I.

To draw a right line, CD, parallel to a given right line, AB, at any given distance, as at

the point D.

With a pair of compasses take the nearest distance between the point D and the given right line, AB; with that distance set one foot of the compasses any where on the line AB, as at A, and draw the arc C on the same side of the line AB as the point D; from the point D draw a line so as just to touch the arc C, and it is done; for the line CD will be parallel to the line AB, and at the distance of the point given, D, as was required.

PROBLEM II.

To bisect or divide a given line, AB, into two equal parts.

Take any distance in your compasses greater than half the line AB; then, with one foot in B, describe the arc CFD; with the same distance, and one foot in A, describe the arc CGD, cutting the former arc in C and D; draw the line CD, and it will bisect AB in the point E.

PROBLEM III.

A

To erect a perpendicular, BA, on the end of a given right line, DB. Take any extent in your compasses, and with one foot in B fix the other in any point, C, without the given line; then, with one point of the compasses in C, describe, with the other, the circle ABD; through D and C draw the diameter DCA, meeting the circle in A; join B and A, and it is done; for BA will be the required line (by Art. 42, Geometry).

Or thus ;

Take any convenient distance, as BH, in your compasses, and, with one foot in B, describe the arc HFG, upon which set off the same distance as a chord from H to F, and from F to G, upon F and G, describe two arcs intersecting each other in A; draw a line from B to A, and it is done; for BA will be the perpendicular required.

PROBLEM IV.

B

B

From a given point, as C, to let fall a perpendicular, CO, on a given right line, AB.

Take any extent in your compasses greater than the least distance between C and the given line AB; with one foot in C, describe an arc to cut the given line, AB, in F and G; with one foot in G, describe an arc, and with the same distance, and one foot in F, describe another arc cutting the former in D; from C to D draw the line COD, cutting AB in Ŏ; then CO will be the perpendicular required.

PROBLEM V.

C

F

B

From a given point, C, to let fall a perpendicular, CB, on a given line, AB, when the perpendicular is to fall so near the end of the given line that it cannot be done as above.

Upon any point, A, of the line AB as a centre, and with the distance AC, describe an arc, E; choose any other point in the line AB, as D, and with the distance DC describe another arc intersecting the former in E; join CE cutting AB in B, and it is done; for CB will be the perpendicular required.

C

B

PROBLEM. VI.

To make an angle that shall contain any proposed number of degrees, from a given point in a given line.

CASE 1. When the given angle is right, or contains 90°, let CA be the given line, and C the given point.

On C erect a perpendicular, CD, and it is done; for the angle DCA is an angle of 90°. Or thus; on the point C, as a centre, with the chord of 60°*, describe an arc, GH, and set off thereon, from G to H, the distance of the chord of 90°, and from C through H draw CHD, which will form the angle DCA of 90° required.`

CASE 2. When the angle is acute, as, for example, 36° 30', let CB be the given line, and C the point at which the angle is to be made.

D

H

90

Α

F

C436.80

B

With the chord of 60° in your compasses, and one foot on C, as a centre, draw the arc FB, on which set off, from B to F, the given angle, 364°, taken from the line of chords; through F and the centre C, draw the right line AC, and it is done; for the angle ACB will be an angle of 36° 30, as was required.

CASE 3. When the given angle is obtuse, as, for example, 127°, let CB be the given line, and C the angular point.

Take the chord of 60° in your compasses, and with one foot on C as a centre, describe an arc, BGHE, upon which set off the chord of 60° (which you already have in your compasses) from B to G, and from G to H; then set off from G to E the excess of the given angle above 60°, which is 67°, taken from the line of chords; or you may set off, from H to E, the excess of the given angle above 120, which is 7°; draw the line CE, and it is done; for the angle ECB will be an angle of 127°.

EH

G

B

Were it required to measure a given angle, the process would have been nearly the same, by sweeping an arc, as BE, and measuring it on the line of chords, as is evident.

PROBLEM VII.

To bisect a given arc of a circle, AB, whose centre is C.

Take in your compasses any extent greater than the half of AB, and, with one foot in A, describe an arc; with the same extent, and one foot in B, describe another arc, cutting the former in D; join CD, and it is done; for this line will bisect the arc AB in the point E. It is also evident that the line CD bisects the angle BCA, or divides it into two equal parts.

PROBLEM VIII.

To find the centre of a given circle.

B

E

E

With any radius, and one foot in the circumference, as at A, describe an arc of a circle, as CBD, cutting the given circle in B; with the same extent, and one foot in B, describe another arc, CAD, cutting the former in C and D; through C and D draw the line CD, which will pass through the centre of the circle; in like manner may another right line be drawn, as EHG, which shall cross the first right line at the centre required. This construction depends upon Art. 43 of Geometry.

PROBLEM IX.

D

To draw a circle through any three given points not situated in a right line. Let A, B, and D, be the given points. Take in your compasses any distance greater than half AB, and, with one foot in A, describe an arc, EF; with the same extent, and one foot in B, describe another arc cutting the former in the points E, F, through which draw the indefinite right line EFC; then take in your compasses any extent greater than half BD, and, with one foot in B, describe an arc, GH; with the same extent, and one foot in D, describe

Tangents

Plate II

another arc cutting the former in the points G, H, through which draw the right line GHC, cutting the former right line EFC in the point C; upon the point C as a centre with an extent equal to CA, CB, or CD, as radius, describe the sought circle.

PROBLEM X.

To divide a circle into 2, 4, 8, 16, or 32 equal parts. Draw a diameter through the centre, dividing the circle into two equal parts; bisect this diameter by another, drawn perpendicular thereto, and the circle will be divided into four equal parts or quadrants; bisect each of these quadrants again by right lines drawn through the centre, and the circle will be divided into eight equal parts; and so you may continue the bisections any number of times. This problem is useful in constructing the mariner's compass.

PROBLEM XI.

To divide a given line into any number of equal parts. Let it be required to divide the line AB into five equal parts. From the point A draw any line, AD, making an angle with the line AB; then through the point B draw a line, BC, parallel to AD; and from A, with any small opening in your compasses, set off a number of equal parts on the line AD, less by one than the proposed number (which number of equal parts in this example is 4); then from B, set off the same number of the same parts on the line BC; then join 4 and 1, 3 and 2, 2 and 3, 1 and 4, and these lines will cut the given line as required. 3

A

C4