and BFD, they are equal in every respect; and, therefore, c 13. 1. DE is equal to DF. In the same manner, it may be shown, that DG is equal to DF; therefore, the three straight lines DE, DF, and DG are equal to one another, and the circle described from the centre D at the distance of any one of them, will pass through the extremities of the other two, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right angles: Therefore, a circle EFG has been inscribed in the given triangle ABC. Which was to be done.

The following problems are so simple, that it is thought unnecessary to subjoin the figures :--

PROP. XIV. PROBLEM.

To inscribe a square in a given circle.

Draw two diameters at right angles to one another, and join their extremities.

PROP. XV. PROBLEM.

To describe a square about a given circle. Through the extremities of two diameters at right angles, draw tangents.

PROP. XVI. PROBLEM.

To inscribe a circle in a given square.

The intersection of the diagonals is the centre, and the perpendicular from it on a side the radius.

Book II.

182

ELEMENTS OF GEOMETRY.

PROP. XVII. PROBLEM.

To describe a circle about a given square.

The intersection of the diagonals is the centre, and its distance from one of the angular points the radius.

PROP. XVIII. PROBLEM.

To inscribe an equilateral and equiangular hexagon in a given circle.

The radius of the circle is the side.

PROP. XIX. PROBLEM.

To describe an equilateral and equiangular hexagon about a given circle.

Divide the circumference into six equal parts, as in the preceding problem, and through the points of section draw tangents.

ELEMENTS
LEME

OF

GEOMETRY.

BOOK III.

DEFINITIONS.

I.-EVERY right-angled parallelogram or rectangle, is Book III. said to be contained by any two of the straight lines which are about one of its angles. Thus, the right-angled parallelogram AC, is called the rectangle contained by AD and DC, or by AD and AB, &c. For the sake of brevity, instead of the rectangle contained by AD and DC, we shall simply say, the rectangle AD.DC, placing a point between the two sides of the rectangle. Also, instead of saying the square of any line, for instance of AD, we shall occasionally write AD2.

These gnomons may also, for the sake of brevity, be called, the first AGK, or EHC, and the second, AKG or HEC.

Book III.

Every quantity is measured by some other quantity of the same kind, of a known magnitude. Thus, a line is measured by a line, a surface by a surface, and a solid by a solid. The measuring unit in the case of lines, may be an inch, a foot, a yard, &c.; in the case of surfaces, a square inch, square foot, square yard, &c.; and, in the case of solids, a cubic inch, cubic foot, cubic yard, &c.

LEMMA.

The product of the two numbers which express the lengths of the adjacent sides of a rectangle, will express the area or quantity of surface contained in it, as measured by the square described upon the linear unit.

Let ABDC be a rectangle, and Ab the linear unit, or unit of length; then, if AB contain this unit 6 times, and AC 3 times, the rectangle will contain a space equal to 18 times the square described upon Ab.

If, through the points of section A
in AB and AC, parallels be
drawn as in the figure, then it
is manifest, that each of the small
surfaces is equal to the square
described on Ab. Now, in the
figure, there are as many squares C
in a row, as there are linear units

し

B

D

in AB, and as many rows as there are linear units in AC;
hence, the whole number of squares will be expressed by the
product of the units in AB by those in AC. In the pre-
sent case, the area will be equal to 6×3=18.
The same

might be shown to hold, although the sides may not be ex-
pressed by whole numbers. Thus, if the length of a rect-
angle were 8 inches, and its breadth 44 inches, then its area
will be equal to 8×4 39 square inches.

=

Cor. The area of a geometrical square is equal to the numerical square of its side. Hence the reason why the product of a number by itself is called the square of the number.

By means of the preceding lemma, the student will be enabled to illustrate the following propositions numerically.

The rectangle contained by two straight lines AC, AB, is equivalent to the sum of the rectangles contained by the one line AC, and the several segments AE, EF, FB, into which the other may be divided.

Or, AC.AB AC.AE+ AC.EF+AC.FB.

Through the points of section E, F, draw EG and FH parallel to AC or BD. AG, EH, and FD rectangles; and AD

=

The figures A
are evidently
AG+EH+

FDa. But AD AC.AB; AG=

=

AC.AE; EH EG.EF = AC.EF, C

=

A

E F B

a Ax. 9.

G II

because EG AC; in like manner, FDAC.FB. Therefore, AC.ABAC.AE+ AC.EF+ AC.FB. Therefore, the rectangle, &c. Q. E. D.

Cor. 1. If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equivalent to the square of the whole line. For, by the Prop. AC.AE+ AC.EF AC.AF; now, suppose AC AF, then we shall have AF.AE+AF.EF = AF2.

=

Cor. 2. If a straight line be divided into any two parts, the rectangle contained by the whole, and one of the parts, is equivalent to the rectangle contained by the two parts, together with the square of the foresaid part. For, as before, AC.AF AC.AE+ AC.EF; and if AC = AE, then AC.AF AE.EF + AE2.

PROP. II. THEOREM.

ไ

If a straight line AB, be divided into any two parts AC, CB, the square of the whole line is equivalent to the squares of the two parts, together with twice the rectangle contained by the parts.

Or, AB2 AC2 + CB2 + 2AC.CB.

=

Upon AB, describe the square ABED, and join BD; through C, draw CGF parallel to AD or BE, and through