In like manner it may be proved that trapezoid DMNC: triangle DCZ =

trapezoid DABC: triangle DCZ

=

AG: BG and

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 not to divide the trapezoid ABCD 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 XLIII. 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 in Z.

2. Upon AB, as a diameter, describe a semicircle, and from Z draw the perpendicular ZM, meeting the semicircle in M.

1

3. Make Ab AM, and upon Ab describe a figure A b c d e, which is similar to the given one ABCDE (see problem XXXIII); the line bc de 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 125); therefore we have the proportion,

ABCDE: Abc de AB X AB : Ab × A b.

=

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

AZ AM: = AM: AB,

and as Ab is by construction equal to AM,

consequently the product A b × Ab is equal to AZ X AB.

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

ABCDE: Abc de AB X AB : AB × AZ, hence ABCDE: Abcde = :AB: AZ; and therefore AB-AZ: AZ;

ABCDE- Abcde: Abcde:

which is read thus:

ABCDE less Abcde is to A b c d e, as AB less AZ is to AZ; that is,

BCDE edc b is to Abc de, 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 XLIV. The three sides of a triangle being given, to construct the triangle.

radius AC, describe an arc of a circle.

B

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

3. 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 XLV. Two sides and the angle included by them being given, to construct the triangle.

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

PROBLEM XLVI. One side and the two adjacent angles being given, to construct the triangle.

SOLUTION. Let AB be the given side, and x

and y, the two adjacent angles.

1. At the two extremities of the line AB, construct the angles x and y, and extend their legs AC, BC, until they meet in the point C; the triangle ABC is the one required.

The demonstration follows from Query 2, Sect. II.

PROBLEM XLVII. Two sides and the angle opposite to the greater of them being given, to construct the triangle.

SOLUTION. Let AC, BC (see the figure to problem XLV) be the two given sides, and the angle, which is opposite to the greater of them (the side BC).

1. Upon an indefinite straight line construct: an angle equal to the angle x.

2. Make the leg AC of this angle equal to the smaller side AC, and from C as a centre, with the radius CB equal to the greater side, describe an arc of a circle, cutting the line AB in the point B.

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

The demonstration follows from Query 10th, Sect. II.

PROBLEM XLVIII. The basis of a triangle, one of the adjacent angles and the height being given, to construct the triangle.