PROBLEM XXXVIII.

To divide a given circle into any proposed number of parts by equal lines, so that those parts shall be mutually equal, both in area and perimeter.

Divide the diameter A B into. the proposed number of equal parts at the points a, b, c, &c. Then on A a, A b, A c, &c. as diameters, describe semicircles on one side of the diameter AB; and on B d, B c, B b, &c. describe semicircles on the other

b

B

side of the diameter. So shall the corresponding joining semicircles divide the given circle in the manner proposed. And in like manner we may proceed, when the spaces are to be in any given proportion. As to the perimeters, they are always equal, whatever may be the proportion o£ the spaces.*

34; and the chords of these equal arcs are equal. The same may be said of each of the other quadrants. Therefore the problem is truly solved.

*The several diameters being in arithmetical progression, the common difference being equal to the least of them, and the diameters of circles being as their circumferences, the circumferences are also in arithmetical progression. But in such a progression the sum of the extremes is equal to the sum of each two terms, equally distant from them; therefore the sum ef the circumferences on AC and CB is equal to the sum of those on AD and DB, and of those on AE and EB, &c. and each sum equal to the semicircumference of the the given circle on the diameter AB. Therefore all the parts have equal perimeters, and each is equal to the circumference of the given circle.

Again the same diameters being as the members 1, 2, 3, 4, &c. and the areas oi circles being as the squares of their diameters, the semicircles will be as the numbers 1, 4, 9, 16, &c. and conse

PROBLEM XXXIX,

On a given Line A B to describe the Segment of a Circle^ capable of containing a given Angle.

To cut off a segment from a given circle, that shall contain a

quently the differences between all the adjacent semicircles are as the terms of the arithmetical progression 1, 3, 5, 7, &c. in which the sums of the extremes, and of every twe equidistant means, constitute the several equal parts of the circle.

* Let full a perpendicular from D upon A B, and it will bisect the angle D. Half the angle D is equal to the complement of the angle DBA the angle ABC= the given angle. But half

1

then DEA will be the segment required, any angle E, made in it, being equal to the given angle C.*

PROBLEM XLl.

To make a triangle similar to a given triangle ABC.

Let ab be one side of the required triangle. Make the angle a equal to the angle A, and the angle b equal to the angle B; then the triangle abc will be similar to ABC, as proposed.

A

B

NOTE. If ab be equal to AB, the triangles will also be 6qual, as well as similar.

PROBLEM KLII.

To make a figure similar to any other given figure ABCDE.

of the circle. Therefore the required segment is described.

For the angle BAD is equal to the angle DEA in the alter

nate segment.

OTHERWISE.

Make the angles at a, b, e, &c. respectively equal to the angles at A, B, E, and the lines will intersect in the angles of the figure required.

PROBLEM XLIII.

To make a triangle equal to a given trapezium ABCD.

Draw the diagonal DB, and CE parallel to it, meeting AB produced in E. Join DE; so shall the triangle ADE be equal to the trapezium AB CD.*

D

C

B

E

PROBLEM XL1V.

To make a triangle equal to the figure ABCDEA

Draw the diagonals DA, DB, and the lines EF, CG, parallel to them, meeting the base AB, both ways produced, in F and G. Join DF, DG; and DFG will be the triangle required.

A

D

B

G

NOTE. Nearly in the same manner may a triangle be made equal to any right-lined figure whatever.

* For the triangles DBE and DBC, being on the same base and between the same parallels, are equal.

PROBLEM XLV.

To make a rectangle, or a parallelogram, equal to a given triangle ABC.

Bisect the base AB in m. Through C draw C n O parallel to AB. Through m and B

draw m n and B O parallel to . each other, and either perpendicular to AB, or making any angle with it. And the rectangle or parallelogram mn O B will be equal to the triangle, as required.*

A

C n

B

PROBLEM XLVI.

To make a square equal to a given rectangle ABCD>

Produce one side AB, till B E be equal to the other side BC. Bisect AE in O; on which as a centre, with radius AO, describe a semicircle, and produce BC to meet it at F. On BF make the

E

B

square BFGH, and it will be equal to the rectangle ABCD, as required.t

*For a parallelogram on half the base and between the same parallels is equal to a triangle.

† For EB=BC: BF :: BF: BA, hence BCxBA=BFs

The given parallelogram and square, that is 'ound, are equal.