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OF THE SYMBOLS EMPLOYED IN THIS TREATISE,
A'S ABBREVIATIONS.

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AB, or AB ...... a straight line, of which the points

AB

AB2

ABX CD

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denoted by A and B are the extremities.

a circular arch, of which the points denoted by A and B are the extremities.

a square, having AB for one of

its sides.

a rectangle, of which AB and CD are adjacent sides.

2AB, &c. ...... the double, &c. of AB. denotes a triangle.

Δ

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.... the ratio of A to B.

A:B::C:D the ratio of A to B is equivalent

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A

SUPPLEMENT

TO THE

ELEMENTS OF EUCLID.

BOOK I.

PROP. I.

1. PROBLEM. A GIVEN plane rectilineal angle being divided into any number of equal angles, to divide the half of it into the same number of angles, all equal to one another.

*

Bisect (E. 9. 1.) the given angle: And, first, if it be divided into an odd number of equal parts, it is evident that the middle part is thereby bisected. Bisect, therefore, each of the remaining

* In this and the following references, the letter E is used to indicate Euclid's Elements; the letter S, in like manner, refers to this Supplement; the former of the subsequent numbers points out the Proposition, and the latter the Book, intended to be quoted.

B

equal parts, on either side of that middle part, and the half of the given angle will, manifestly, be divided into as many equal parts as the given angle itself.

Again, if the given angle be divided into an even number of equal parts, it is plain that the straight line which bisects it, will have the half of that number of equal parts, on each side of it. Bisect, therefore, each of the equal parts, on either side of that line; and the half of the given angle will thereby be divided, as before, into as many equal parts as the given angle itself.

2. PROBLEM.

PROP. II.

From the vertex of a given scalene triangle, to draw, to the base, a straight line which shall exceed the less of the two sides, as much as it is itself exceeded by the greater.

Let ABC be the given scalene triangle, and let AB be greater than AC: It is required to draw,

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from the vertex A, to the base BC, a straight line which shall exceed AC, as much as it is exceeded by AB.

From AB cut off (E. 3. 1.) AD=AC; bisect (E. 10. 1.) DB in E; from the centre A, at the distance AE, describe (E. 3. Post.) the circle EF cutting BC in F; and join (E. 1. Post.) A, F: Then is AF the straight line which was to be drawn.

=

For, (E. 15. def. 1.) AFAE; and (constr.) AD AC; ... AF-ACAE-ADDE. Also, AB-AE BE; i. e. AB-AF BE: and (constr.) BE—DE.

... AF-AC-AB-AF.

PROP. III.

3. PROBLEM. In a straight line given in position, but indefinite in length, to find a point, which shall be equidistant from each of two given points, either on contrary sides, or both on the same side of the given line, and in the same plane with it; but not situated in a perpendicular to it.

Let XY be a given straight line indefinite in length, and A, B, two given points without it; not situated in a perpendicular to XY: It is required to find a point in XY that shall be equidistant from A and B.

First, let A, B be both on the same side of XY:

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