and there is sufficient room on both sides of the given line (AB). With the centre A (Fig. 9), and any radius greater than the half of AB, describe an are; with the centre B, and the same radius, describe another arc, cutting the former in D and E; join the points of intersection D and E, cutting the line AB in C, which will bisect AB in the point C.

B

3RD METHOD. When the given line (AB) occurs near the bottom of the drawing, and sufficient room is only left on one side of the given line (AB). With centres A and B (Fig. 10), and any equal radii greater than the half of AB (this radius should be as long as can be conveniently got), describe arcs cutting each other in the point D. With the same centres, and less equal radii, also greater than the half of AB, describe arcs cutting each other in the point E; join DE, and produce the line until it cuts AB in C, which will be the point of bisection of AB.

Fig. 9.

PROBLEM II.

To draw a perpendicular or a straight line at right angles to

a given straight line (AB) from a given point (C). A (Euclid, Book I., Prop. XI. and XII.)

1ST METHOD.-When the given point (C) is within or without the given line (AB) and near the middle of it. With the centre C (Fig. 11), and

any radius, describe an arc cutting the line AB in the
points E and F.
With the centres EF, and the
same, or any other equal radii, describe arcs cutting
each other in the point D; join DC, then the line
DC is perpendicular to the the line AB as required.

Fig. 12.

-B

-

2ND METHOD.
When the given point
(C) is within and at or
near the extremity of
the given line (AB).
Take any point F
(Fig. 12) above the

line AB, and with the
radius FC describe

the arc ECD, cutting

AB in the point E; join EF, and produce it until it cuts the arc ECD in the point D, and join DC, which will be perpendicular to AB, as required.

3RD METHOD.—When the given point (C) is without and opposite the extremity of the given line (AB). In

Fig. 11.

the line AB (Fig. 13) take any two convenient points E, F. With the centre E, and radius

EC, describe the arc CGD, and with the centre F, and radius FC, describe an arc cutting the former arc CGD in the point C and D ; join CD; the line CD will be perpendicular to the given line AB, as required.

F

B

A very simple method of setting off a perpendicular on the ground may be performed with a tape line in the following manner :-Suppose it was required to set out a straight line of road (CD) (Fig. 12) at right angles, or A perpendicular to another straight line of road (AB) from any point (C) in AB. Put in a pin at the point C, and also one at E, along one side of the line of road AB, and three feet distant from C; put the end of the tape line at E, pass it round the pin at C, carrying it up in the direction of the line of new road, and bringing twelve feet on the line to the pin at E, and holding twelve feet and the end of the tape line fast at E, stretch the line tight, putting in another pin at seven feet on the line. When stretched, this will give the point D, and form a triangle, having sides three, four, and five feet respectively in length, of which DC will be perpendicular to EC, and form one side of the new road required.

Fig. 13.

This method of setting off a perpendicular is taken from that beautiful theorem in Euclid (Book I., Prop. XLVII.), where in this case ED, the hypothenuse of the triangle, is equal to 5 feet, and the two sides EC and CD, 3 and 4 feet respectively; therefore 52 = 32 + 42, or 25 916. The sum of the three sides being 12, or the whole length of line used; but any numbers may be used (consistent with the length of the tape line), provided they fulfil the required value. It may be easily remembered that any multiple of the numbers used, viz., 3, 4, 5, will suffice.

=

1ST METHOD.-Take any two points, D and E A

D

Fig. 14.

E

B

(Fig. 14), in the given line AB, the further apart from each other the better, with the centres D and E and radii equal to the distance between the point C and the line AB ;* describe two arcs,

and draw a straight line to touch or be a tangent to both arcs, then the line GH is parallel to AB.

2ND METHOD. From the point C (Fig.15) draw the line CD perpendicular to AB; at another point anywhere in the line AB, but as far from D as convenient, draw E H perpendicular to A B. Make EH equal to CD, and draw through the points C and H the straight line FG, which will be parallel to AB.

C

F

H

G

-B

Fig. 15.

*The distance between C and the line AB is easily got by placing one point of the compasses at C, and opening them until the other point just touches, or describes an arc, AB. FG may be drawn parallel to AB at any distance apart, by merely using that distance as the radii of the arcs.

E

C

H

-G

-B

D

3RD METHOD.-Take any point D (Fig.16), the further from C the better, in the line AB; join DC, and with the centre D, and radius DC (or any other convenient radius which will cut DC), describe an arc cutting AB in E; also with the centre C, and the same radius, describe an arc DH; again with the centre D, and a radius equal to the distance between C and E, describe an arc cutting the latter arc in HI; draw the line FG passing through the points C and H; the line FG will be parallel to AB. 4TH METHOD. Take any point D (Fig. 17) in the line AB; join DC and produce. it, making CO equal to CD; with the centre O and radii OC and OD, describe two arcs, the latter cutting AB in E; join OE, cutting the lesser arc in the point H; draw the line FG passing through the points C and H, then FG will be parallel to AB.

Fig. 16.

5TH METHOD.—Take any point D (Fig. 18) in the line AB; join DC and produce it; with the centre C, and radius CD, describe a circle, cutting AB in E, and DC produced in 0; with the centres E and O, and the same radius, describe arcs intersecting in the point H; draw FG passing through the points C and H; then FG will be parallel to AB.

PROBLEM IV.

To divide a given straight line (AB) into any given number (say 6) of equal parts. Through A (Fig. 19) draw a line AD at any angle with AB, and through B draw

BC parallel to AD; from A set off six equal parts. on the line AD (those equal parts should be as nearly the length of the required equal parts of AB as can be estimated), and from B on the line BC set off the same number (six) equal parts of the same length as before; join B and o, 1 and 1,2 and 2, 3 and 3, 4 and 4, 5 and 5, A and 6; then the points where those lines cut AB, viz., in a, b, c, d, e, will divide the line AB into the required number (six) f equal parts.

А

5

4

3

Fig. 19...

B.

PROBLEM V.

B

To find a mean or mean proportional between two given lines (AB and BC). Draw a straight line ABC (Fig. 20) equal in length to AB and BC together; bisect AC in 0,* and with the centre O and radius OA, or OC, describe a semicircle; from B draw a line perpendicular to AC, and meeting the semicircle in the point D, then BD is a mean or mean proportional between AB and BC.

E

D

To describe a square on a given straight line (AB). (Euclid, Book I., Prop. XLVI.) From the point A (Fig 21) draw AE perpendicular to AB (Problem II.), and from AE cut off AD equal to AB; draw DC and CB parallel to AB and AD respectively, intersecting at the point C; or, with B and D as centres, and radii equal to AB, describe arcs intersecting at the point C, and join DC and BC, then ABCD is the square required.*

To describe a square which shall be equal to any number of given squares (1, 2, 3, 4, 5).

Draw any two lines, AB and AC

(Fig. 22), at right angles to each other; in AB make AD equal to the side of

the square 1, and in AC make AE equal to the side of the square 2, and join DE; then in AB make AF equal to DE, and in AC make AG equal to the side of the square 3, and join FG; again, in AB make AH equal to FG, and in AC AK equal to the side of the square 4, and join HK; lastly, in AB make AL equal to HK, and in AC make AM equal to the side of the square 5, and join LM; then the square LM NO, described on the line LM, is equal to the sum of the squares 1, 2, 3, 4, 5.

By proceeding in a similar manner till all the sides of any number of given squares are employed, a square may be constructed equal to the sum of the given number of squares ; this, as well as the next problem, depending throughout their whole construction on (Euclid, Book I., Prop. XLVII.)

* Of course any rectangle whose length and breadth are given can be constructed in a similar manner, by making the perpendicular equal to the breadth, and either drawing parallels or using the length of the sides as radii.

PROBLEM VIII.

To describe a square which shall be equal to the difference of any two given squares (F and G).

Draw any line GK, which shall be a mean proportional between the two sides, AB and BC of the given parallelogram (Problem V.), and upon the line GK describe the square HGKL, which will be equal to the given parallelogram, ABCD. (Fig. 24.)

When the given parallelogram is not a rectangle, as AEFB, then draw a rectangle ABCD, standing on the

D

E

с

T

same base, and between the same parallels (Euclid, Book I., Prop. XXXV.), equal to it, and proceed as before.

PROBLEM X.

To draw through a given point (F) a straight line, which shall tend to the intersection of two given straight lines (AB, CD), but whose point of intersection (0) falls beyond the limits of the drawing (LMN).

E

H

1ST METHOD.-Through the given point E, draw any line EC, meeting AB in A, and CD in C, draw any other line IIK parallel to EC, meeting AB and CD in G and K; join CG, and through K draw KF parallel to it, meeting AB in F; join EG, and through F draw FH parallel to it, meeting KG produced in H; join EH, which, if produced, would pass through the point of intersection 0.

F B

D

N

M

Fig. 25.