PROBLEM II.

To find a fourth proportional to three given lines, A, B, C.

Draw the two indefinite lines DE, DF, forming any angle with each other. Upon DE take DA A, and DB=B; upon DF take DC=C; draw AC; and through the point B, draw BX

parallel to AC; DX will be the fourth proportional required; for, since BX is parallel to AC, we have the proportion DA: DB:: DC: DX; now the first three terms of this proportion are equal to the three given lines: consequently DX is the fourth proportional required.

Cor. A third proportional to two given lines A, B, may be found in the same manner, for it will be the same as a fourth proportional to the three lines A, B, B.

PROBLEM III.

To find a mean proportional between two given lines A and B.

Upon the indefinite line DF, take DE A, and EF-B; upon the whole line DF, as a diameter, describe the semicircle DGF; at the point E, erect upon the diameter the perpendicular EG meeting the circumference in G; EG will be the mean proportional required.

D

E

BH

AH

For, the perpendicular EG, let fall from a point in the circumference upon the diameter, is a mean proportional between DE, EF, the two segments of the diameter (Prop. XXIII. Cor.); and these segments are equal to the given lines A and B.

PROBLEM IV.

To divide a given line into two parts, such that the greater part shall be a mean proportional between the whole line and the other part.

Let AB be the given line. At the extremity B of the line AB, erect the perpendicular BC equal to the half of AB; from the point C, as a centre, with the radius CB, describe a semicircle; draw AC cutting the circumference in D; and take AF-AD:

F

the line AB will be divided at the point F in the manner required; that is, we shall have AB : AF :: AF: FB.

For, AB being perpendicular to the radius at its extremity, is a tangent; and if AC be produced till it again meets the circumference in E, we shall have AE: AB:: AB : AD (Prop. XXX.); hence, by division, AE-AB : AB :: ABAD AD. But since the radius is the half of AB, the diameter DE is equal to AB, and consequently AE-AB=AD=AF; also, because AF-AD, we have AB-AD=FB; hence AF: AB : FB : AD or AF; whence, by exchanging the extremes for the means, AB : AF :: AF: FB.

Scholium. This sort of division of the line AB is called division in extreme and mean ratio: the use of it will be perceived in a future part of the work. It may further be observed, that the secant AE is divided in extreme and mean ratio at the point D; for, since AB-DE, we have AE: DE :: DE: AD.

PROBLEM V.

Through a given point, in a given angle, to draw a line so that the segments comprehended between the point and the two sides of the angle, shall be equal.

Let BCD be the given angle, and A the given point.
Through the point A, draw AE paral-

lel to CD, make BE-CE, and through
the points B and A draw BAD; this will
be the line required.

For, AE being parallel to CD, we have BE: EC :: BA: AD; but BE=EC; therefore BA=AD.

E

D

B

A

PROBLEM VI.

To describe a square that shall be equivalent to a given parallelogram, or to a given triangle.

First. Let ABCD be the given parallelogram, AB its base, DE its altitude: between AB and DE find a mean proportional XY; then will the square described upon

D

A E

B

XY be equivalent to the parallelogram ABCD.

For, by construction, AB XY:: XY: DE; therefore, XY2=AB.DE; but AB.DE is the measure of the parallelogram, and XY2 that of the square; consequently, they are equivalent.

Secondly. Let ABC be the given triangle, BC its base, AD its altitude: find a mean proportional between BC and the half of AD, and let XY be that mean; the square described upon XY will be équivalent to the triangle ABC.

For, since BC: XY:: XY AD, it follows that XY2= BC. AD; hence the square described upon XY is equivalent to the triangle ABC.

PROBLEM VII.

Upon

a given line, to describe a rectangle that shall be equivalent to a given rectangle.

Let AD be the line, and ABFC the given rectangle.

Find a fourth propor

tional to the three lines

C

X

F

tional; a rectangle constructed with the lines A AD and AX will be equi

valent to the rectangle ABFC.

B

For, since AD : AB : : AC: AX, it follows that AD.AX= AB.AC; hence the rectangle ADEX is equivalent to the rectangle ABFC.

PROBLEM VIII.

To find two lines whose ratio shall be the same as the ratio of two rectangles contained by given lines.

Let A.B, C.D, be the rectangles contained by the given lines A, B, C, and D.

Find X, a fourth proportional to the three lines B, C, D; then will the two lines A and X have the same ratio to each other as the rectangles A.B and C.D.

For, since B: C:: D: X, it follows that C.D=B.X; hence A.B: C.D:: A.B: B.X :: A : X.

Cor. Hence to obtain the ratio of the squares described upon the given lines A and C, find a third proportional X to the lines A and C, so that A: C::C: X; you will then have

A.X-C2, or A2.X-A.C2; hence
A2 C2 A : X,

PROBLEM IX.

To find a triangle that shall be equivalent to a given polygon.

Let ABCDE be the given polygon. Draw first the diagonal CE cutting off the triangle CDE; through the point D, draw DF parallel to CE, and meeting AE produced; draw CF: the polygon ABCDE will be equivalent to the polygon ABCF, which has one side less than the original polygon.

G

B

A E F

For, the triangles CDE, CFE, have the base CE common, they have also the same altitude, since their vertices D and F, are situated in a line DF parallel to the base: these triangles are therefore equivalent (Prop. II. Cor. 2.). Add to each of them the figure ABCE, and there will result the polygon ABCDE, equivalent to the polygon ABCF.

The angle B may in like manner be cut off, by substituting for the triangle ABC the equivalent triangle AGC, and thus the pentagon ABCDE will be changed into an equivalent triangle GCF.

The same process may be applied to every other figure; for, by successively diminishing the number of its sides, one being retrenched at each step of the process, the equivalent triangle will at last be found.

Scholium. We have already seen that every triangle may be changed into an equivalent square (Prob. VI.); and thus a square may always be found equivalent to a given rectilineal figure, which operation is called squaring the rectilineal figure, or finding the quadrature of it.

The problem of the quadrature of the circle, consists in finding a square equivalent to a circle whose diameter is given.

PROBLEM X.

To find the side of a square which shall be equivalent to the sum or the difference of two given squares.

F

A

B

E

EG B; draw DG: this will be the side of the square required.

For the triangle DEG being right angled, the square described upon DG is equivalent to the sum of the squares upon ED and EG.

Secondly. If it is required to find a square equivalent to the difference of the given squares, form in the same manner the right angle FEH; take GE equal to the shorter of the sides A and B; from the point G as a centre, with a radius GH, equal to the other side, describe an arc cutting EH in H: the square described upon EH will be equivalent to the difference of the squares described upon the lines A and B.

For the triangle GEH is right angled, the hypothenuse GH-A, and the side GE=B; hence the square described upon EH, is equivalent to the difference of the squares A and B.

sum

Scholium. A square may thus be found, equivalent to the of any number of squares; for a similar construction which reduces two of them to one, will reduce three of them to two, and these two to one, and so of others. It would be the same, if any of the squares were to be subtracted from the sum of the others.