of which are together equal to two right angles, a circle may be circumscribed about it, that is, the circumference which passes through the three points A, B, C, shall also pass through D.

A

D

For if D were to lie within the circle, the angle D would be greater than if it were in the circumference (Prop. XVI. Cor. 1.), and consequently the angles B, D would be together greater than two right angles (Prop. XVII.); and if D were to lie without the circle the angle D would be less than if it were in the circumference, and therefore the angles B, D would, in this case, be less than two right angles; D therefore can lie neither within nor without the circle, that is, it is in the circumference.

B

Cor. 1. If two opposite angles of a quadrilateral be together equal to the other two opposite angles, a circle may be described about the quadrilateral (Prop. XVII. Cor. 3. B. I.).

Cor. 2. A trapezium may be inscribed in a circle, provided the non-parallel sides are equal.

D

For if the non-parallel sides AD, BC, of the trapezium ABCD, be equal, and perpendiculars DE, CF, be drawn to AB, the triangles ADE, BCF will be equal, the angle A A E equal to the angle B, and the angle

F B

ADE to the angle BCF, and consequently the angle ADC must be equal to the angle BCD, so that two opposite angles of the trapezium are equal to the other two, therefore a circle may be described about it (Cor. 1.).

BOOK IV.

POSTULATE.

From any point as a centre with any radius, a circumference may be described.

PROPOSITION I. PROBLEM.

To divide a given straight line AB, into two equal

parts.

From the points A and B as centres, with any radius greater than half AB, describe two arcs, which must necessarily cut each other (Prop. XIII. Schol. 4. B. III.); draw the straight A line CD through the points of intersection, and it will pass through M, the middle of AB; for CD is perpendicular to AB (Prop. XIII. Cor. 3. B. III.),

DC

and the points A, B, are equally distant from C, therefore they are equally distant from M (Prop. XXII. Schol. B. I.).

Cor. CD not only divides AB into two equal parts, but it is at the same time perpendicular to AB.

From a given point P, in a straight line AB, to draw a perpendicular to that line.

In the given straight line, or in its prolongation, take two points C D, equally distant from P; and from these points as centres, with a radius longer than CP, describe arcs which will intersect in E; draw PE and it will be the perpendicular required; for it is drawn

A

-B

P

D

from the middle of the straight line CD to a point equally distant from its extremities (Prop. XXII. Cor. 4. B. I.).

Scholium.

If the point P were the extremity of the line, and if the line could not be produced beyond it, then a different construction must be employed. Thus:

From any point C taken without the line, with a radius equal to the distance CP, describe a circumference, and from D, where it cuts AP, or its prolongation, draw the diameter DE; then EP will be the perpendicular required, as is manifest from Prop. XIV. Cor. 3. B. AIII.

From a given point P, without a straight line AB, to draw a perpendicular to that line.

Take any point C in AB, and from P

as a centre, with a radius equal to the
distance PC, describe an arc CD, and
from the points C, D, as centres, with
the same, or any other radius, describe AC
two arcs cutting in E, then PE will be
the perpendicular required, for it passes
through two points P, E, each of which
is equally distant from the two extremities
of CD (Prop. XXII. Cor. 5, B. I.).

Scholium.

P

-B

If the point P were opposite the extremity of the line AB, or nearly so, and if AB could not be produced beyond this extremity, the following construction may be employed.

From the other extremity of AB, or from any other point in AB, with a radius equal to its distance from P, describe an arc; then from a second point in AB, with its distance from P as a radius, describe another arc, and through their points of intersection draw a line, which will be the perpendicular required; as is obvious from (Prop. XIII. Cor. 3. B. III.).

A

PROPOSITION IV. PROBLEM.

At a point P, in a straight line AB, to make an angle equal to a given angle D.

From P as a centre,

with any radius, de

scribe an arc CE, and

from D as a centre with

the same radius, de

scribe an arc FG, ter- D minating in the sides of

the angle D; then, with a radius equal to the chord of this arc, describe, from the point C as a centre, an arc cutting the arc CE in H; draw HP, and HPC will be equal to the angle D; for in equal circles equal arcs subtend equal angles at the centre (Prop. IV. B. IÏI).

PROPOSITION V. PROBLEM.

To divide a given angle, ACB, into two equal parts. From C as a centre, with any radius, describe an arc AEB, terminating in the sides of the angle, and draw CD perpendicular to its chord; then the angle ACB will be bisected (Prop. V. B. III.).

Scholium.

A

B

E

By repeated bisections an angle may obviously be divided into four, eight, sixteen, &c. equal parts, but the division of an angle into three equal parts is a problem that cannot be generally effected by elementary geometry. This problem is one of very ancient date, and for a long time engaged the attention of some of the greatest geometers of Greece, who were at length compelled to relinquish the hope of performing this operation by a purely geometrical method, that is to say, by employing no other lines in the construction of the problem

* The trisection of a right angle, and hence of,,, &c. thereof, is a very easy problem, the construction of which may be left for the student to perform. It may not be improper to remark here, that an ingenious instrument, for the mechanical trisection of angles, has recently been devised by Mr. R. Christie, for a description of which see Mechanic's Magazine, vol. I.

than the straight line and the circumference of a circle. By the introduction of other curves, the trisection has been effected in various ways.

PROPOSITION VI. PROBLEM.

Two angles of a triangle being given, to find the third angle.

D

Draw any straight line AB, and take therein a point P, at which make an angle APC equal to one of the given c angles, and then another CPD, equal to the other given angle: the third angle DPB will be equal to the third angle of the triangle, for the three angles of the triangle are together equal Ato the three angles at the point P, each amounting to two right angles, and two of the angles at P have been made equal to two angles of the triangle, therefore the third angle must be equal to the remaining angle of the triangle.

P

B

PROPOSITION VII. PROBLEM.

Two angles and a side of a triangle being given, to construct the triangle.

First, let the angles be adjacent to the given side. Draw the straight line BC, equal to the given side, and at the extremities make two angles, BCA, CBA, equal to those given, then the sides BA, CA, must meet and form with BC the triangle required; for if they were parallel, the angles B, C, would be together equal to two right angles (Prop. XIV. Cor. 4. B. I.), and therefore could not belong to a triangle.

B

But if one of the given angles be opposite to the given side, then find the other angle by last proposition, and proceed as above.

PROPOSITION VIII. PROBLEM.

Two sides of a triangle, and the angle which they include being given, to construct the triangle.

Draw BC (preceding diagram) equal to the given side, make