A man and a boy are similar figures; a house and a well-constructed model of it are similar figures; or, a machine and a model of it.

Any two squares are similar; any two equilateral triangles are similar; any two circles are similar.

172. Similar rectilineal figures have the angles of one equal, each to each, to the angles of the other, and the corresponding sides, about the equal angles, in the same proportion; that is, any two sides in one triangle are in the same proportion as the corresponding sides (173) in the other triangle.

B

FIG. 44

Да

D

The triangles ABC, DEF, are similar. Their angles are equal, each to each-namely, AD; B = E; C=F; and the sides about the equal angles are in the same proportion; those opposite equal angles being the antecedents, or consequents of the ratios—

173. AB and AC are corresponding sides to DE and DF. They are the sides of equal angles, A and D. AB corresponds to DE, being opposite equal angles, C and F. AC and DF correspond; they are opposite the equal angles B and E. To say that sides correspond, means that they are similarly situated in respect to equal angles. AB and DE, opposite the equal angles C and F, are the antecedents of the ratios;

DIVISION OF A LINE.

51

AC and DF, opposite the equal angles B and E, are the consequents of the ratios.

Problem 42.

174. To divide a straight line into any number of equal parts.

Let AB be the line, to be divided into five equal parts.

FIG. 45

3

Through A and B, draw AC and BD parallel to each other, on opposite sides of AB.

From A towards C, lay off on AC four equal partsone less than the number given-A 1, 1-2, 2-3, 3-4; and the same number of the same length each, from B towards D, on BD.

Join 4, the last point marked on AC, to 1, the first marked on BD, 3 to 2, 2 to 3, and 1 to 4. These cross lines will divide the line AB into five equal parts.

175. Note. This may be done in a simpler manner, but requiring a greater number of lines to be drawn parallel to each other.

Lay off five equal parts on any line, AF, making an

angle with AB,-namely, A 1, 1-2, 2-3, 3-4, 4-F. Join BF. Through the four points of division, draw lines parallel to BF. These will divide AB into five equal parts.

176. Note.-This illustrates the geometrical truth, that

Parallel lines cut diverging lines in the same proportion.

The points, 1, 2, 3, 4, divide the line AF into five equal parts, and the parallel lines drawn through these points divide the line AB in the same proportion; that is, into five equal parts.

Problem 43.

177. To divide a straight line into any number of equal parts, without drawing parallel lines.

Let it be desired to divide AB into five (or n) equal parts.

Draw any line AC, making an acute angle with AB.

On it lay off six (n + 1) equal parts. Let E be the end of the 4th (n-1th) part, reckoning from A. Join CB, and produce it till the produced part (BD) is equal to CB. Join DE.

From B to the point F, where DE cuts AB, will be one-fifth of AB.

178. To find a fourth proportional to three given straight lines.

Let AB, BC, and AD be the lines; it is required to find a fourth line, x, such that—

AB BC: AD: x.

Place AB and BC in the same straight line, and AD making an angle with AC. Join BD, and through C, draw CE parallel to BD, meeting AD produced in E. DE is the line required; that isAB: BC:: AD: DE.

B

FIG. 47

E

Note. This illustrates the above geometrical truth (176). BD parallel to CE, cuts AC and AE in the same proportion.

Problem 45.

179. To find a third proportional to two given straight lines, AB, BC, last figure; that is, a line, x, such thatAB: BC: BC: x.

Draw AD

Place AB and BC in one straight line. equal to BC, and making any angle with AC. Join BD. Through C draw CE parallel to BD. DE is the line required; that is, AB : BC :: BC (or AD): DE.

Problem 46.

180. To find a mean proportional between two given straight lines, AB and BC, fig. 40, page 45; that is, to find a straight line, x, such that

AB x:: x: BC.

This is done in Problem 35, page 45.

Proceed exactly

as is done in that problem. BE is the line required. AB: BE:: BE: BC.

Note.-If EA, EC, fig. 40, were joined, AEC would be a triangle, right-angled at E (72). Whence it appears that if from the right angle of a right-angled triangle a perpendicular be drawn to the hypotenuse, the perpendicular is a mean proportional between the segments of the hypotenuse. Farther, such perpendicular divides the triangle into two triangles similar to the large triangle and to each other.

Problem 47.

181. To divide a straight line, so that one part is to the other as the latter is to the whole line.

Let AB be the line; it is required to divide it into two parts, Ax and xB, such that—

Ax: xB

FIG. 48

xB : AB.

At A draw AC perpendicular to AB and equal to half of it. Join св. From C, with radius CA, describe an arc cutting CB in D. From B, with radius BD, describe an arc, cutting AB

The point E divides AB as required; that is

AE EBEB: AB.

Problem 48.

182. To produce a given line, so that the whole line, as produced, shall be to the given line as the latter is to the produced part.

Let AB be the given line (fig. 49).

It is required to produce it by the additional line, BC, such that

AC: AB:: AB: BC.

Bisect AB in D. Through B draw BE perpendicular