gles u and n are opposite angles at the vertex, and w and s are alternate angles (page 31, 2d); therefore the triangle BtH, having the side Bt, and the two adjacent angles, w and u, equal to the side tE, and the two adjacent angles, n and s, in the triangle GtE, these two triangles are equal to one another; consequently the area of the trapezoid ABGH (composed of the quadrilateral ABGt, and the triangle BtH) is equal to the area of the triangle ABE (composed of the same quadrilateral ABGt and the equal triangle GtE), which proves the correctness of construction 3.

PROBLEM XL. To divide a trapezoid into a given number of equal parts, so that the lines of division may be parallel to the parallel sides of that trapezoid. [This problem may be omitted by the younger pupils.]

SOLUTION. Let ABCD be the given trapezoid which is to be divided into three equal parts.

1. Upon AB, the greater of the two parallel sides, describe a semicircle; draw DE parallel to CB; and from B, with the radius BE, describe the arc of a circle, EF, cutting the semicircle in F.

2. From F draw FG perpendicular to AB, and divide

the part AG of the line AB, into three equal parts in K and I; from these points draw the perpendiculars Kk, Ii.

3. Upon AB, from B towards A, take the distances Bm, Bn, equal to Bk, Bi; from the points m and n, draw the lines mO, nM, parallel to BC; and from the points O, M, in which these parallels meet the side AD, the lines MN, OP, parallel to AB; then ABNM, MNPO, OPCD, are the three required parts of the trapezoid ABCD

DEMON. Extend the lines AD, BC, until they meet in Z. Then the triangles DCZ, OPZ, MNZ, ABZ, are all similar to each other (page 70); further, we have (by construction 3)

The areas of the two similar triangles OPZ, CDZ, are in the ratio of the squares upon the corresponding sides; that is, we have the proportion

triangle OPZ: triangle DCZ: OP X OP: CD X CD; and since OP is equal to Bk, and CD to BF,

also

triangle OPZ: triangle DCZ= Bk × Bk : BF × BF. Imagine AF and FB joined; the triangle AFB would be rightangled in F, and we should have the proportion

BG: BFBF: AB;

and for the same reason we have

BK: Bk=Bk: AB.

Taking the product of the mean and extreme terms of the two last proportions, we have

BG X AB equal to BF X BF, and

Let us now take our first proportion,

triangle OPZ: triangle DCZ= Bk × Bk : BF × BF;

and let us write BG X AB, instead of BF × BF (its equal), and BK X AB, instead of Bk × Bk, and we shall have

triangle OPZ: triangle DCZAB X BK: ABX BG,

whence

triangle OPZ: triangle DCZ= BK: BG;

consequently, also,

triangle OPZ — triangle DCZ : triangle DCZ=BK — BG : GB; (Principle 6th of Geom. Prop. page 62); which is read thus: triangle OPZ, less the triangle DCZ, is to the triangle DCZ,

as the line BK, less the line BG, is to the line BG; that is, trapezoid DOPC: triangle DCZ= GK : BG; and as GK is (by construction 2) equal to of AG,

trapezoid DOPC : triangle DCZ=} AG: BG.

In like manner it may be proved that

trapezoid DMNC: triangle DCZ=} AG: BG and
trapezoid DABC: triangle DCZ=AG: BG.

These proportions express that the three trapezoids DOPC, DMNC, DABC, are to each other in the same proportion as one third is to two thirds to three thirds; or, which is the same, as one is to two, to three; whence the rest of the demonstration follows of course.

Remark. If it is required to divide the trapezoid ABCD not into equal parts, but according to a given proportion, it will only be necessary to divide the line AG in this proportion, and then proceed as before.

PROBLEM XLI. To divide a given figure into two parts according to a given proportion, and in such a way, that one of the parts may be similar to the whole figure.

SOLUTION. Let ABCDE be the given figure.

1. Divide one side of the figure, say AB, according to the given proportion; let the point of division be Z.

2. Upon AB, as a diameter, describe a semicircle, and

from Z draw the perpendicular ZM, meeting the semicircle in M.

3. Make Ab AM, and upon Ab describe a figure, Abcde, which is similar to the given one, ABCDE (see Problem XXXIII); the line bcde divides the figure in the manner required.

DEMON. The areas of the two similar figures Abcde, ABCDE, are to each other, as the squares upon their corresponding sides (page 98); therefore we have the proportion

ABCDE: Abcde AB X AB : Ab × Ab.

Draw AM and BM; then AM is a mean proportional between AZ and AB; that is, we have

AZ: AMAM: AB;

and as Ab is, by construction, equal to AM,

AZ: Ab=Ab: AB;

consequently the product Ab × Ab is equal to AZ × AB.

Writing AZ X AB, instead of Ab × Ab (its equal), in the first proportion, we have

=

ABCDE: Abcde AB X AB: AB X AZ. Hence ABCDE: Abcde=AB: AZ; and therefore ABCDE-Abcde: Abcde=AB-AZ: AZ; which is read thus:

ABCDE, less Abcde, is to Abcde as AB, less AZ, is to AZ; that is,

BCDEedcb is to Abcde as ZB is to AZ;

consequently the figure ABCDE is divided according to the given proportion in which the line AB is divided.

PART IV.

Construction of triangles.

PROBLEM XLII.

The three sides of a triangle being

given, to construct the triangle.

SOLUTION. Let AB, AC, BC, be the three given sides of the triangle.

1. Take any side, say AB, and from A as a centre, with the radius AC, describe

an arc of a circle.

ААВ

C

B

2. From B, as a centre, with the radius BC, describe another arc, cutting the first.

4. From the point of intersection C, draw the straight lines CA, CB; the triangle ABC is the one required.

The demonstration follows immediately from Query 4th, Sect. II.

PROBLEM XLIII. Two sides, and the angle included by them, being given, to construct the triangle.

SOLUTION. Let AB, AC, be the two given sides, and

x the angle included by them.

1. Construct an angle equal to the angle x (Problem VI); make one of the legs equal to the side AB, and the other to the side AC.

2. Join BC; the triangle ABC is the one required.

The demonstration follows from Query 1, Sect. II.