this meets the circumference draw ST and S'T' perpendicular to PP'; these will be parallel to AD (P. 19, B. 1) and tangent to the circumference OP (P. 8).

There are two solutions.

PROBLEM XIV.

To draw a tangent to a given circumference from a point without the circumference.

N

Let P be a point without the circumference OM. Draw PO, and on it, as a diameter, construct a circumference cutting the given circumference at M and N; then draw PM and PN; they will pass through P and be tangent to the circumference OM. For draw the radii OM and ON; the angles PMO and PNO are right angles because they are inscribed in semicircles; hence, PM and PN are perpendicular to the radii OM, and ON, at M, and N; they are therefore tangent to the circumference OM; the problem has two solutions.

Cor. The right-angled triangles PNO and PMO have a common hypothenuse PO, and the angles OPN, and OPM are equal, because they are measured by halves of equal arcs ON and OM; the triangles are therefore equal in all their parts (P. 17, B. 1); hence, the tangents PN and PM are equal, and the line PO bisects the angle between them, also the angle NOM, between the radii to the point of contact.

PROBLEM XV.

To inscribe a circle in a given triangle.

Let ACD be the given triangle. Bisect the angles A and C, and prolong their bisectrices till they meet at O. The point

M

A

O is equally distant from the three sides of the triangle. For draw OM, ON, and OP, respectively perpendicular to AC, CD, and DA; because AO is the bisectrix of the angle A, OM and OP are equal, and because CO is the bisec- c trix of the angle C, OM and ON are equal hence, the three perpendiculars are all equal. centre, with OM as a radius, describe a circumference and it will be tangent to all the sides of the given triangle; hence, the circle OM is inscribed in the triangle ACD.

PROBLEM XVI.

(P. 18, B. 1);

To circumscribe a circle about a given triangle. Let ACD be the given triangle. Bisect AC and CD by perpendiculars (Prob. 1), and prolong them till they meet at 0; from O as a centre with the radius OA describe a circumference and it will pass A through all the vertices of the given tri

From O as a

angle (P. 7); hence, the circle OA is circumscribed about the given triangle.

PROBLEM XVII.

To construct a segment on a given line that shall contain a given angle.

Let AC be the given line and Q the given angle. At A construct the angle CAT equal to Q, and draw AM perpendicular to AT at A; bisect AC by a perpendicular and prolong it till it meets AM at O; with O

as a centre and OA as a radius, draw a circumference; it will pass through C and be tangent to AT at A. Then will any angle, as ADC, inscribed in the segment CDA be equal to Q; for the angles CDA and CAT are each measured by half the arc AC (P. 14, and P. 14, Cor. 4); but the angle CAT is equal to Q; hence, CDA is also equal to Q.

ADDITIONAL PROBLEMS.

Let the pupil solve the following additional problems, using the annexed diagrams as guides to their solution.

PROB. XIX. Through a given point P draw a circumference tangent to a given line CB at a given point B.

PROB. XXI. Draw a circumference tangent to a given line TP and to a given circumference CQ at a given point Q.

PROB. XXII. Draw a circumference with a given radius tangent to a given line DP and to a given circumference CT.

PROB. XXIII. From two given points A and C draw two lines meeting on a given line BD and equal to each other.

PROB. XXIV. Inscribe a circle in a given quadrant BCDP.

PROB. XXV. Draw a circumference through a given point P within a given angle CBA that shall be tangent to both sides of the angle.

B

BOOK III.

RATIOS, PROPORTIONS, AND COMPARISON OF LINES.

Definitions and Principles.

65. The ratio of one magnitude to another of the same kind is the quotient of the second by the first. The first magnitude is called the antecedent and the second is called the consequent.

The operation of dividing one magnitude by another consists in dividing the number of times that any assumed unit is contained in the former by the number of times it is contained in the latter. If the magnitudes are commensurable the ratio is exact; if not, it is only approximate.

Let P and Q be two incommensurable magnitudes of the same kind. Suppose P to be divided into m equal parts, m being any whole number, and let Q contain n of these parts with a remainder less than one of them; then will,

The greater the value of m, the more nearly will

proach to the true value of the ratio of P to Q.

66. The limit of a varying magnitude, is a value towards which that magnitude may approach, with which it may coincide, but beyond which it cannot pass.

It is assumed that whatever is true of all values of a varying magnitude is true of its limit.