half AF, D is the middle of AH. Hence the semicircle passes through H, AGH in the semicircle is a right angle *.

and

Scholium. In practice these methods are sometimes superseded by others analogous to those spoken of in the last Scholium, especially where A is not very remote from the line BC. The former method is more convenient than the latter of those processes.

Adjust the scale so that the cross line CF shall coincide with AB, and the side DE shall pass through A; and draw AC by the edge DE.

Or, thus again, by the scale.

Place the ruler abcd along the given line, and the shorter side of the triangle efg against the ruler: then the longer one will be perpendicular to AB, however it be slid along the ruler. If the point A be not more distant from BC, the triangle in being slid along may be made to pass through A, and the line drawn along it will be the perpendicular required. If, on the contrary, the side fg will not reach A', draw any perpendicular AH by means of it, and then through A', a line A'H' parallel to this perpendicular. It will be that required.

A very accurate rectangular ruler may be made by doubling a piece of stiff paper, ABCD, so as to obtain a straight line EF; and then carefully doubling again in GH, so that GF shall coincide with GE.

A

B a

C

PROBLEM VII.

To bisect a given straight line AB.

First method. With centres A, B, and any convenient equal radii, describe circles intersecting in C and D. Join CD: then it will bisect AB in E.

For it may be shown, as before, that CE bisects the vertical angle CB of the isosceles triangle, and hence also it bisects the base (th. 3. Cor. 1) ↑.

Second method. From one extremity A draw any line AG, and take any equal distances AC, CD, in it. Join DB, and draw CE parallel to it, meeting AB in E, then E is the middle of AB.

For the line CE being parallel to the base DB of the

E

E

* This method is generally given in the form of " take any line AH and bisect it in D," and so on. The present only differs as to the mode of finding the middle point D. The process is applicable, in this latter form, to the case where G falls near the corner of the drawing.

It is not essential that the radii of the circles intersecting in C and of those in D should be equal; but when circumstances admit, it is convenient to take them so as the compasses require no alteration. They need not even be on different sides of the line AB; and hence, when AB is near an edge of the picture, it will be requisite to take these pairs of radii different, and obtain two intersections C and D on the same side of AB.

triangle BAD, it divides the sides AD, AB, proportionally. But AD is bisected in C; hence AB is bisected in E *.

Third method. From the two extremities A, B, draw parallel lines on alternate sides of AB. Take AD, BC, any equal lines, and join CD: it will bisect AB in E.

For, obviously, the triangles are equal, having the angles at A and D equal to those at B and C, each to each, and

D

the side AD equal BC. Whence the sides AE, EB, are also equal †.

B

Scholium. When the common parallel ruler is used, a still better form of construction will be:

From A, B, draw any two pairs of parallels intersecting in C and D; then the diagonal CD will bisect the diagonal AB.

Fourth method. Draw any line KL parallel to AB, and taking any point D in it, set off CD, DF, on each side of any equal lengths. Join CA, FB, meeting in G or CB, FA, meeting in G': draw the line DG or DG', cutting AB in E. Then E is the middle of AB.

For by similar triangles, AE: EB:: CD: DF, and CD DF; hence AE = EB. In the same way

for the point G′ ‡.

Fifth method. Draw any line AF, and in it take any three equal distances AC, CD, DH; join HB, and produce it till BG = BH; join CG, cutting AB in E. This is the middle of AB.

For, (th. 95,) AE : EB :: GH. AC: BG. CH :: 2BG. AC: 2AC. BG. Hence, since the last two terms are equal, the two former are so too: that is, AE = EB, and AB is bisected in E.

D

C

B

E

G

F

B

L

PROBLEM VIII.

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

First method. Draw any straight line AK making an angle with AB, and in it set off AC, CD, DE.......... all equal to one another (and as nearly as possible of the estimated length of one of the required parts of AB as can be estimated, will be convenient): join FB, and draw EN, DM, CL, ... parallel to FB. The sections of these lines with AB will be the points sought.

C

D

* This method is convenient when a parallel ruler is used, and AB is near and parallel to a margin of the drawing.

This method is also convenient where AB is not nearly parallel and adjacent to a margin, if at the same time a parallel ruler be used.

This method is very convenient when there is a line KL parallel to AB already in the drawing.

For by parallels, AC: CD: DE .... :: AL: LM: MN .... and as the former are all equal, the latter will be so too *.

G

A

B

M

L

...

Second method. Take any two parallel lines AC, BD; and in them respectively set off equal parts AE, EF, FG .... and BH, HI, IK .....; the number of parts in each being one less than the number of parts into which AB is to be divided. Draw EK, FI, GH ..., (the point E being the nearest to A, and K the most remote from B,) meeting AB in L, M, N .... Then these will be the points of division sought. For by the parallels AC, BD, and the construction, EK, FI, GH parallel, and hence the line AB cutting them is divided proportionally †. Third method. Draw any line KL parallel to AB, and in it take CD, DE, EF, FG, GH ... all equal to one another, and the same number as there are to be of parts formed of AB. Join AC, HB, the extremes meeting in P; and draw PD, PE, PF, PG .... meeting AB in M, N, R, S.... Then these will be the points of division sought.

For by the parallels, CD: DE: EF ........:: AM : MN: NR .... and the former being all equal, the latter are so too .

....

Fourth method. Draw any line AC making an angle (a small one better than a large one) with AB. Set off equal distances AX, XV ........ FE, ED, such that AE is as many times AX as AB is of the part required, and let D and F be the points of division one on each side of E adjacent to it. Join DB, and produce it till BH = BD; join FH, cutting AB in G: then BG is the part of AB required.

For, (th. 95,) AF. DH. BG = FD. HB. GA. Hence, if we denote by m the number of parts into

K C DE

MV

DEFGH

which AB is to be divided, we have AE= m. AX, AF = (m − 1) AX, FD = 2AX, and DH = 2DB. Whence (m—1) AX. 2BH. BG = 2AX. BH.

AG or

AG= (m 1) GB; and therefore AG + GB = mGB: that is,
AB = mGB, or GB is the mth part of AB §.

* This is a convenient method in practice if the parallel ruler or parallel scales are used. It, however, can also be used when AB is near and nearly parallel to the margin of the drawing. + This is a convenient process when the compasses and ruler are alone used. In that case, the lines AC, BD, should be the sides of the equilateral triangles described on AB, one on each side. It requires, however, that AB should not be near and nearly parallel to the margin of the picture.

When the line KL can be taken sufficiently long, it will be desirable to take it at least about twice the length of AB, as otherwise the point P will fall very remote from AB. When this cannot be done, and the line will allow of sufficiently distinct divisions, take it about or less than half the length of AB, as in the lower position of the figure, for the same reason. When it is possible to take it at some distance from AB, set off the divisions nearly equal to the estimated divisions of AB, and draw AH, CB, crossing between AB and KL, as in the corresponding construction of the last problem.

§ This process is due to M. Chenou, Professor of Mathematics in the Royal College of Douay,

A

D

E

K

M

NL H F

Fifth method. On AB describe any parallelogram ABCD, and draw the diagonals AC, BD, intersecting in E; draw EF parallel to AD, meeting AB in F, and join DF, meeting AC in G; draw GH parallel to AD, meeting AB in H, and join DH meeting AC in K. Continue this process as far as may be necessary for the purpose. Then AF is the half of AB, AH is the third of AB, AL is the fourth of AB; and so on, as far as the operations may be carried on. Since BD is bisected in E (th. 22), and EF is parallel to the base AD of the triangle ABD, we have AF

=

FB.

Since AD: EF :: AB: BF, we have AD

2EF, and hence by parallels we have also AF: AH :: AE: AG :: DF: DG :: 3 : 2, or 2AF: AH :: 6 : 2, or again AB = 3AH.

In a similar manner may the truth of the succeeding divisions be proved: but it is more fully detailed in the demonstration of Prob. 7 of the next section on Practical Geometry in the Field.

PROBLEM IX.

D

To draw a third proportional to two given lines AC, BD. TAKE any two lines EF, EG, meeting in E; and in these respectively take EH = AB, and EK EI = BD. Draw KL parallel to IH: then EL is a third proportional to AB and CD.

For by parallels EH: EI :: EK = EI): EL; and by construction, EH = AB, EI = EK = BD; whence AC: BD :: BD: EL *.

H

PROBLEM X.

To draw a fourth proportional to three given lines AB, CD, EF.

TAKE two lines GH, GI, meeting in G, and in GH take GK, GM, equal to the first and third lines AB, EF, and in GI take GL equal to the second line CD. Draw MN parallel to KL, meeting GI in N. Then GN is the fourth proportional to the three given lines AB, CD, EF. The proof is similar to that of the last proposition ↑.

A C

G

K

м

H

I

and is much used by the French draughtsmen. Its chief recommendation is, that it dispenses with the use of parallels; whilst its chief objection arises from the time and care required to get the compasses accurately set to the distance BG, so as to slip off the other points of division of the line AB, the smallest amount of error being so multiplied in the process as to create a great difference between the several intermediate portions and the last towards A. It is well adapted to the case, where only the nth part of a given line is required without the other points of the line AB being sought; but where all are required, it is inferior in precision and simplicity, and the preceding methods are preferable, especially if parallel instruments be allowed.

* The solution may be varied in several different but very obvious ways; but they are all alike in principle.

The same remark applies to this as to the last problem.

PROBLEM XI.

B

To find a mean proportional between two given lines AB, BC. First method. Let them be placed in one straight line, as in the figure: on AC, as diameter, describe a semicircle, and from B draw BD perpendicular to AC, meeting the circle in D. Then BD is the mean proportional required.

This is evident from th. 87 *.

Second method. Let them be placed in a straight line as before. With any equal radii from centres A and C describe arcs intersecting in E and F, and let O be the intersection of EF, AC. Make BP = BO, and from O and P as centres, with radii equal to AO describe arcs intersecting in D. Then DB will be the mean proportional between AB and BC.

For the first part of the construction finds O the middle of AB, and the second at the same time finds BD perpen

dicular to AC and OD = OA. The construction is therefore identical with the preceding one.

Scholium. The fifth method of bisecting the line AC also works well with the subsequent operations here employed.

Third method. Let AB, BC, be placed as in the figure. Bisect AC in O, and describe semicircles on AC and BO, intersecting in H. Then BH is the mean proportional required.

For by the construction, the angle OHB is in

H

a semicircle, and hence is a right angle, and AOH a radius of the circle AHC. Whence BH is a tangent to AHC, and hence AB: BH :: BH: BC †.

PROBLEM XII.

To find any number of continued proportionals to two given lines AB, AC.

ON AC, the greater of the two lines, describe a semicircle, in which place the less line AB, and produce both lines indefinitely. Draw CD perpendicular to AC, meeting AB produced in D; draw DE perpendicular to AD, meeting AC produced in E; draw EF perpendicular to AE, meeting AD produced in F; and Then AB, AC, AD, AE.... are a series of continued proportionals.

so on.

*This method implies more actual work than the statement of it might lead us to expect. The next method, which is in reality identical with this, comprises the entire operation.

+ This method, though more laborious, is sometimes convenient: viz. when one or both the given lines are of considerable length. Its advantage in this case is that the circles do not require so much space, nor are they so difficult to describe, as in the other processes.