Sol. At the point B in the line BC make an angle ABC equal to FED, and at the point C make an angle ACB=EFD; then are triangles ABC and EFD similar, (264.)

271. To determine the distance AB, (fig. 144,) which cannot be directly measured.

Sol. Mark out a straight line AC, and take in it any point D. Measure angle A, and at D make the angle CDE=A. Produce DE until it meets a supposed line in the direction CB at the point E. We shall thus have triangle CDE. Since angle A = D therefore DE is parallel to AB; therefore CD: CA=DE: AB.

The three lines CD, CA, and DE may be measured; and therefore we can find AB. For example, if CD = CA, then DE=1 AB. Take the length of DE four times and we have AB.

272. To prepare a scale of equal parts, (fig. 145.) This is a scale upon which straight lines or lengths are divided into a certain number of equal parts. It is used in drawing upon paper representations in a reduced size of lines and figures. Suppose it is required to draw a plan of a field having five sides; that is, to construct a pentagon similar to the field. It would not be convenient to construct upon paper a pentagon whose sides should be equal to those of the field; neither is it necessary so to do; we have only to make the angles upon paper equal to the corresponding angles in the field, and to draw the sides upon paper in the same ratio to each other as the sides of the field. Any length may be taken to represent a rod or a foot. Suppose one side of a field is 10 feet 9 inches in length, and that the diagram is to be made upon a scale of one foot to an inch; that is, one inch on paper is to be taken as equal to one foot in the field. Measure between the points of the compasses a distance equal to 10 inches, and this will represent the 10 feet, then if the inches on the scale are divided into 12 equal parts, extend the opening of the compasses by a distance equal to 9 of these parts, for each of the parts represents one inch. Draw upon the paper an indefinite line, and take in it a part equal to the opening of the compasses. The use of a scale of equal parts is to enable us to construct small figures similar to great figures. To construct a scale of equal parts, some one unit of length, for example an inch, is taken as the basis, and several of these units are marked upon the rule, or the paper on which the scale is to be made. One of the extreme divisions is again divided into many equal parts. The unit assumed as the basis is generally small, and great accuracy is necessary in the division; if the division was made in the common mode, the marks would be so near one to another as not easily to be distinguished. A mode of division founded on the proportion of the sides of similar triangles has been adopted. In a straight line AB, (fig. 145,) take the parts AC, CD, DB, each equal to the unit of length assumed as the basis of the scale. Divide AC into 10 equal parts; then if AC=1 inch, each of these parts will equal to of an inch. Now to get the hundredths of an inch, construct upon AC any rectangular figure, for example the square AFMC. Divide each of the sides into 10 equal parts. Through the points of division of the lines AF and CM draw straight lines parallel to AC. Connect the division points of the lines AC and FM by diagonal lines; that is, connect 10 in one line with 9 in the other; 9 in one line with 8 in the other, &c. By the construction MN =AC,

therefore MN: OP=MC : OC=10:9; that is, OP = MN,

MN: QR=MC:QC=10:8;

that is, QR = MN.

In like manner it may be demonstrated that UV has 7 of the parts of which MN has 10. Thus the extent MN is divided into 10 equal parts, and any one of these parts may be taken between the points of the compasses with the greatest ease. If AC=100 feet, and we have a line 255 feet in length to be marked off; then extend the points of the compasses from 2 of the larger divisions to 5 on the 5th line of the subdivisions, that is on line GH.

Let us suppose a field ABCDE, (fig. 157,) is to be represented upon paper by a scale of 100 rods to an inch. We must first measure each of the sides and angles of the field. Suppose the side AB is 100 rods long; draw the line a b one inch long, and it will represent the side AB. At b make, with a protractor, an angle equal to angle B, and this will give us the direction of the next side b c, which, if BC is 120 rods in length, will be 1 inch and of an inch. At c make an angle equal to C, and we shall thus get the direction of the next side. Proceed in this manner till the construction of the figure is completed The figure abcde will be an accurate representation of the field; for, by construction, the angles of the one are equal to the homologous angles of the other; and the sides of the one are proportional to the homologous sides of the other. Therefore the two are similar figures.

273. To find the perpendicular height BC of a tower, when it cannot be directly measured, (fig. 146.)

Sol. 1. Measure upon the base of the tower the horizontal line AB. Draw upon paper the line ED representing AB upon a reduced scale. Measure the angle BAC, and make upon the paper the angle EDF= BAC. Erect at E the perpendicular EF. The triangle DEF is similar to ABC, (264,) because by construction angle EDF = BAC, and DEF = ABC;

therefore DE: EF = AB : BC.

But DE, EF, and AB are known quantities, therefore BC may be known. If DE= EF, then AB=BC and BC = AB; suppose AB=60 feet; then BC = X6080 feet.

Sol. 2. Set up in the clear sunlight a perpendicular staff EF, (fig. 146.) Measure the length of the shadow ED which it casts. Measure also the length of the shadow which the tower makes at the same time. The rays of the sun's light may be considered as parallel; therefore, these rays, that is, the lines CA and FD make equal angles with the horizontal lines BA and ED; that is, angle CAB FDE. Now angle CBA = FED, being right angles; therefore triangle CBAS FDE,

therefore DE: EF=AB: BC, the height of the tower. That is, the length of the shadow of the staff is to the height of the staff as the length of the shadow of the tower is to the height of the

tower.

274. To find from the windows of a house the height of an object, (fig. 147.)

Let A and B be the windows, one of which is perpendicularly over the other, and CG the height to be found. Sol. Measure from A, the angle CAE, which an imaginary line CA drawn from the top of the object makes with the imaginary horizontal line AE. Measure from B the angle CBD, made by the imaginary line CB with the imaginary horizontal line BD.

Now angle

CAB=CAE+EAB=CAE +1 R. A. CAE+90°;

and angle

ABC=ABD-CBD=1 R. A. — CBD=90° — CBD. Now we know the angles CAE and CBD, and therefore CAB and ABC may also be known; thus in the triangle ABC, the side AB and the two adjacent angles are known; therefore we can construct a similar triangle; and thus find the length of BC. We shall then know in the triangle BCD, the side BC and the angles CBD and CDB, (which is a right angle.) By means of these we can find the length of CD. Again, measure from B the angle DBG, made by DB with an imaginary line GB, drawn from the bottom of the object to the point B. In the triangle DBG the side DB and its adjacent angles are known; we can therefore find the length of DG. Add DG to CD and we shall have the whole height required.

275. To find the height of an object which cannot be approached very nearly; for example CD, (fig. 148.)

Sol. In the direction AD, measure the line AB. Measure also the angle CAB and CBD. By means of the scale of equal parts draw upon paper the line EF to represent AB. Make the angle FEG = CAB. Produce FE and make the angle GFH=CBD. From the point G let fall the perpendicular GH.

Triangle EFGS ABG, therefore we have the proportions,

EF: FG AB: BC,

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