To construct a Parallelogram. To find the Centre of a Circle.

sides, and, by § 128, find the point C at the given distances AC and BC from the point C, join AC and BC, and ABC is the triangle required.

144. Scholium. The problem is impossible, when one of the given sides is greater then the sum of the other two.

145. Problem. To construct a right triangle, when a leg and the hypothenuse are given.

Solution. Draw AB (fig. 79) equal to the given leg. At A erect the perpendicular AC, from B as a centre, with a radius equal to the given hypothenuse, describe an arc cutting AC at C. Join BC, and ABC is the triangle required.

146. Problem. The adjacent sides of a parallelogram and their included angle being given, to construct the parallelogram.

Solution. Make the angle A (fig. 80) equal to the given angle, take AB and AC equal to the given sides, find the point D, by § 128, at a distance from B equal to AC, and at a distance from C equal to AB. Join BD and DC, and ABCD is, by § 80, the parallelogram required.

147. Corollary. If the given angle is a right angle, the figure is a rectangle; and, if the adjacent sides are also equal, the figure is a square.

148. Problem. To find the centre of a given circle or of a given arc.

Solution. Take at pleasure three points A, B, C (fig. 81) on the given circumference or arc; join the chords AB and BC, and bisect them by the perpendiculars DE and FG; the point O in which these perpendiculars meet is the centre required.

To draw a Tangent to a Circle.

Proof. For, by § 117, the perpendicular DE and FG must both pass through the centre, which must therefore be at their point of meeting.

149. Scholium. By the same construction a circle may be found, the circumference of which passes through three given points not in the same straight line, or in which a given triangle is inscribed.

150. Problem. Through a given point, to draw a tangent to a given circle.

Solution. a. If the given point A (fig. 82) is in the circumference, draw the radius CA, and through A draw AD perpendicular to CA, and AD is, by § 120, the tangent required.

b. If the given point A (fig. 83) is without the circle, join it to the centre by the line AC; upon AC as a diameter describe the circumference AMCN, cutting the given circumference in M and N; join AM and AN, and they are the tangents required.

Proof. For the angles AMC and ANC are right angles, because they are inscribed in semicircles, and therefore AM and AN are perpendicular to the radii MC and NC at their extremities, and are, consequently, tangents, by 120.

151. Corollary. The two tangents AM and AN are equal; for the right triangles AMC and ANC are equal, by 64, since they have the hypothenuse AC common, and the leg MC equal to the leg NC, and, therefore, the other legs AM and AN are equal.

152. Problem. To inscribe a circle in a given triangle ABC (fig. 84).

Solution. Bisect the angles A and B by the lines AO

To inscribe a Circle in a Triangle.

and BO, and their point of intersection is the centre of the required circle, and the perpendicular OD let fall from O upon the side AC is its radius.

OAE

Proof. The perpendiculars OD, OE, and OF let fall from O upon the sides of the triangle are equal to each other. For in the right triangles OAD and OAE the hypothenuse OA is common; the angle OAD by construction; and the third angle AOD §67; the triangles are, therefore, equal, by § 53 ; and OD is equal to OE. In the same way it may be proved that OF OD OE.

AOE, by

Hence the circumference DFE passes through the points D, F, E, and the sides are tangents to it, by § 120.

153. Corollary. The three lines AO, BO, and CO, which bisect the three angles of a triangle, meet at the same point.

154. Problem. Upon a given straight line AB (figs. 85 and 86), to describe a segment capable of containing a given angle, that is, a segment such that each of the angles inscribed in it is equal to a given angle.

Solution. Draw BF, making the angle ABF equal to the given angle. Draw BO perpendicular to BF, and OC perpendicular to the middle of AB. From O, the point of intersection of OB and OC, with a radius OB OA, describe the circumference BMAN, and BMA is the segment required.

=

Proof. Since BF is perpendicular to BO, it is a tangent to the circle, and therefore the angles AMB and ABF are equal, since they are each, by § 107 and 121, measured by half the arc ANB.

155. Scholium.

If the given angle were a right angle,

To find the Ratio of two Lines.

the segment sought would be a semicircle described upon the diameter AB.

156. Problem.

To find a common measure of two given straight lines, AB, CD (fig. 87), in order to express their ratio in numbers.

Solution. a. The method of finding the common divisor is the same as that given in arithmetic for two numbers. Apply the smaller CD to the greater AB, as many times as it will admit of; for example, twice with a remainder BE.

Apply the remainder BE to the line CD, as many times as it will admit of; twice, for example, with a remainder DF.

Apply the second remainder DF to the first BE, as many times as it will admit of; once, for example, with a remainder BG.

Apply the third remainder BG to the second DF, as many times as it will admit of.

Proceed thus till a remainder arises, which is exactly contained a certain number of times in the preceding.

This last remainder is a common measure of the two proposed lines; and, by regarding it as unity, the values of the preceding remainders are easily found, and, at length, those of the proposed lines from which their ratio in numbers is deduced.

If, for example, we find that GB is contained exactly three times in FD, GB is a common measure of the two proposed lines.

CD=2. EB + FD=8+3=11,

AB 2. CD+ EB 22+4=26;

AB=2.

To divide a Line into equal Parts.

consequently, the ratio of the lines AB, 11; that is, AB is {{ of CD, and CD is

CD is as 26 to

of AB.

157. Corollary. By a like process, may be found the ratio of any two quantities, which can be successively applied to each other, like straight lines, as, for instance, two arcs or two angles.

CHAPTER X.

PROPORTIONAL LINES.

158. Theorem. If lines a a', b b', c c', &c. (fig. 88), are drawn through two sides AB, AC of a triangle ABC, parallel to the third side BC, so as to divide one of these sides AB into equal parts A a, a b, &c., the other side AC is also divided into equal parts Aa', a'b', &c.

Proof. Through the points a', b', c', &c. draw the lines a' m, b'n, c'o, &c. parallel to AB.

The triangles Aaa', a'm b', b'n' c, &c. are equal, by § 53; for the sides a' m, b' n, c' o, &c. are, by § 79, respectively equal to ab, bc, cd, &c., and are therefore equal to each other and to Aa; moreover, the angles Aaa', ma' b', nb' c', &c. are equal, by § 29, and likewise the angles Aaa', a' m b', b'n c', &c. Consequently, the sides A a', a' b', b' c', &c. are equal.

159. Problem. To divide a given straight line AB (fig. 89) into any number of equal parts.