PROBLEM.

149. To find the centre of a given circle, or of a given arc.

Solution. Take at pleasure three points A, B, C (fig. 84), in Fig. 84. the circumference of the circle or in the given arc; join AB and BC, and bisect them by the perpendiculars DE, FG; the point O, in which these perpendiculars meet, is the centre sought.

150. Scholium. By the same construction a circle may be found, the circumference of which will pass through three given points A, B, C, or in which a given triangle ABC may be inscribed.

PROBLEM.

151. Through a given point, to draw a tangent to a given circle. Solution. If the given point A (fig. 85) be in the circumfer- Fig. 85, ence, draw the radius CA, and through A draw AD perpendicular to CA, then AD will be the tangent sought (110). If the point A (fig. S6) be without the circle, join the point A and the centre Fig 86. by the straight line AC; bisect AC in O, and from O, as a centre, with the radius OC, describe an arc cutting the given circle in the point B; draw AB, and AB will be the tangent required.

For, if we draw CB, the angle CBA inscribed in a semicircle is a right angle (128); therefore AB, being a perpendicular at the extremity of the radius CB, is a tangent.

152. Scholium. The point A being without the circle, it is evident that there are always two equal tangents AB, AD, which pass through the point A; they are equal (56) because the right-angled triangles CBA, CDA, have the hypothenuse CA common, and the side CB = CD; therefore AD = AB, and at the same time the angle CAD = CAB.

PROBLEM.

153. To inscribe a circle in a given triangle ABC (fig. 87). Fig. 87. Bisect the angles A and B of the triangle by the straight lines AO and BO, which will meet each other in O; from the point O draw the perpendiculars OD, OE, OF, to the three sides of the triangle; these lines will be equal to each other. For, by construction, the angle DAO = OAF, and the right angle ADO= AFO ;

Fig. 88,

89.

consequently the third angle AOD is equal to the third AOF. Besides, the side AO is common to the two triangles AOD, AOF ; therefore, a side and the adjacent angles of the one being respectively equal to a side and the adjacent angles of the other, the two triangles are equal; hence DOOF. It may be shown, in like manner, that the two triangles BOD, BOE, are equal; consequently OD = OE; therefore the three perpendiculars OD, OE, OF, are equal to each other.

Now, if from the point O, as a centre, and with the radius OD, we describe a circle, it is evident that this circle will be inscribed in the triangle ABC; for the side AB, perpendicular to the radius at its extremity, is a tangent. The same may be said of the sides BC, AC.

154. Scholium. The three lines, which bisect the three angles of a triangle, meet in the same point.

PROBLEM.

155. Upon a given straight line AB (fig. 88, 89) to describe a segment capable of containing a given angle C, that is a segment such, that each of the angles, which may be inscribed in it, shall be equal to a given angle C.

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Solution. Produce AB toward D, make at the point B the angle DBE C, draw BO perpendicular to BE, and GO perpendicular to AB, G being the middle of AB; from the point of meeting O, as a centre, and with the radius OB, describe a circle; the segment required will be AMB.

For, since BF is perpendicular to the radius at its extremity, BF is a tangent, and the angle ABF has for its measure the half of the arc AKB (131); besides, the angle AMB, as an inscribed angle, has also for its measure the half of the arc AKB; consequently the angle AMB = ABF = EBD = C ; therefore each of the angles inscribed in the segment AMB is equal to the given angle C.

156. Scholium. If the given angle were a right angle, the segment sought would be a semicircle described upon the diameter AB.

PROBLEM.

157. To find the numerical ratio of two given straight lines AB, CD (fig. 90), provided, however, these two lines have a common Fig. 90.

measure.

Solution. 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, for example, once 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 will be the 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 twice in FD, GB will be the common measure of the two proposed lines. Let GB = 1, we have FD = 2; but EB contains FD once plus GB; therefore EB=3; CD contains EB once plus FD; therefore CD 5; AB contains CD twice plus EB; therefore AB = 13; consequently the ratio of the two lines AB, CD, is as 13 to 5. If the line CD be considered as unity, the line AB would be ; and, if the line AB be considered as unity, the line CD would be.

158. Scholium. The method, now explained, is the same as that given in arithmetic for finding the common divisor of two numbers (Arith. 61), and does not require another demonstration.

It is possible, that, however far we continue the operation, we may never arrive at a remainder, which shall be exactly contained a certain number of times in the preceding. In this case the two lines have no common measure, and they are said to be incommensurable. We shall see, hereafter, an example of this in the ratio of the diagonal to the side of a square. But, although the exact ratio cannot be found in numbers, by neglecting the last remainder we may find an approximate ratio to a greater Geom.

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or less degree of exactness, according as the operation is more or less extended.

PROBLEM.

Fig, 91.

159. Two angles A and B (fig. 91) being given, to find their common measure, if they have one, and from this their ratio in

numbers.

Solution. Describe, with equal radii, the arcs CD, EF, which may be regarded as the measure of these angles; in order then to compare the arc CD, EF, proceed as in the preceding problem; for an arc may be applied to an arc of the same radius, as a straight line is applied to a straight line. We shall thus obtain a common measure of the arcs CD, EF, if they have one, and their ratio in numbers. This ratio will be the same as that of the given angles (122); if DO is the common measure of the arcs, DAO will be the common measure of the angles.

160. Scholium. We may thus find the absolute value of an angle by comparing the arc, which serves as its measure, with the whole circumference. If, for example, the arc CD is to the circumference as 3 to 25, the angle A will be of four right angles, or of one right angle.

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It may happen, as we have seen with respect to straight lines, that arcs also, which are compared, have not a common meas. ure; we can then obtain, for the angles, only an approximate ratio in numbers, more or less exact, according to the degree to which the operation is extended.

SECTION THIRD.

Of the Proportions of Figures.

DEFINITIONS.

161. I SHALL call those figures equivalent whose surfaces are equal.

Two figures may be equivalent, however dissimilar; thus a circle may be equivalent to a square, a triangle to a rectangle, &c.

The denomination of equal figures will be restricted to those, which being applied, the one to the other, coincide entirely; thus two circles having the same radius are equal; and two triangles

having the three sides of the one equal to the three sides of the other, each to each, are also equal.

162. Two figures are similar, which have the angles of the one equal to the angles of the other, each to each, and the homologous sides proportional. By homologous sides are to be understood those, which have the same position in the two figures, or which are adjacent to equal angles. The angles, which are equal in the two figures, are called homologous angles. Equal figures are always similar, but similar figures may be very unequal.

163. In two different circles, similar arcs, similar sectors, similar segments, are such as correspond to equal angles at the centre. Thus, the angle A (fig. 92) being equal to the angle O, the Fig. 92. arc BC is similar to the arc DE, the sector ABC to the sector ODE, &c.

164. The altitude of a parallelogram is the perpendicular which measures the distance between the opposite sides AB, CD (fig. 93), considered as bases.

Fig. 93. The altitude of a triangle is the perpendicular AD (fig. 94), Fig. 94. let fall from the vertex of an angle A to the opposite side taken for a base.

The altitude of a trapezoid is the perpendicular EF (fig. 95) Fig. 95. drawn between its two parallel sides AB, CD.

165. The area and the surface of a figure are terms nearly synonymous. Area, however, is more particularly used to denote the superficial extent of the figure considered as measured, or compared with other surfaces.

THEOREM.

166. Parallelograms, which have equal bases and equal altitudes, are equivalent.

Demonstration. Let AB (fig. 96) be the common base of the Fig. 96. two parallelograms ABCD, ABEF; since they are supposed to have the same altitude, the sides DC, FE, opposite to the bases, will be situated in a line parallel to AB (69). Now, by the nature of a parallelogram, AD= BC (84), and AF = BE; for the same reason, DC AB, and FE AB; therefore DC=FE. If DC be taken from DE, there will remain CE; and if FE, equal to DC, be taken also from DE, there will remain DF ; consequently CE = DF.

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