Scholium. The same construction serves for making a circumference pass through three given points A, B, C'; and also for describing a circumference, which shall circumscribe a given triangle ABC. PROBLEM XIV. Through a given point, to draw a tangent to a given circle. Let A be the given point, and C the centre of the given circle. If the given point A lies in the circumference, draw the radius CA, and erect AD perpendicular to it: AD will be the tangent required (P. 9). If the point A lies without the circle, join A and the centre, by the straight line CA: bisect CA in 0; from O as a centre, with the radius OC, describe a circumference intersecting the given circumference in B; draw AB: this will be the tangent required. For, drawing CB, the angle CBA being inscribed in a semicircle is a right angle (P. 18, c. 2); therefore, AB is a perpendicular at the extremity of the radius CB; hence, it is a tangent (P. 9). Scholium 1. When the point A lies without the circle, there will be two equal tangents, AB, AD, passing through the point A for, there will be two right-angled triangles, CBA, CDA, having the hypothenuse CA common, and the side CB= CD; hence, there will be two equal tangents, AB, AD. The angles CAD, CAB, are also equal (B. 1., P. 17). Scholium 2. As there can be but one line bisecting the angle BAD, it follows, that the line which bisects the angle formed by two tangents, must pass through the centre of the circle. PROBLEM XV. To inscribe a circle in a given triangle. B E D Let ABC be the given triangle. Bisect the angles A and B, by the lines AO and BO, meeting in the point 0 (PROB. 5); from the point O, let fall the perpendiculars OD, OE, OF (PROB. 3), on the three sides of the triangle these perpendiculars will all be equal. A F For, by construction, we have the angle DAO=OAF, the right angle ADO=AFO; hence, the third angle AOD is equal to the third AOF (B. I., P. 25, C. 2). Moreover, the side AO is common to the two triangles AOD, AOF; and the angles adjacent to the equal side are equal: hence, the triangles themselves are equal (B. I., P. 6); and DO is equal to OF. In the same manner it may be shown that the two triangles BOD, BOE, are equal; therefore OD is equal to OE; hence, the three perpendiculars ØD, OE, OF, are all equal. Now, if from the point O as a centre, with the radius OD, a circle be described, this circle will be inscribed in the triangle ABC (D. 11); for, the side AB, being perpendicular to the radius at its extremity, is a tangent (P. 9); and the same thing is true of the sides BC, AC. Scholium. The three lines which bisect the three angles of a triangle meet in the same point. PROBLEM XVI. On a given straight line to describe a segment that shall contain a given angle; that is to say, a segment such, that any angle inscribed in it shall be equal to a given angle. Let AB be the given straight line, and C the given angle. Produce AB towards D. At the point B, make the angle DBE=C; draw BO perpendicular to BE, and at the middle point G, draw GO perpendicular to AB: from the point 0, where these perpendiculars meet, as a centre, with the distance OB, describe a circumference: the required segment will be AMB. For, since BF is perpendicular to the radius OB at its extremity, it is a tangent (P. 9), and the angle ABF is measured by half the arc AKB (P. 21). Also, the angle AMB, being an inscribed angle, is measured by half the are AKB (P. 18): hence, we have AMB ABF=EBD=C: hence, any angle inscribed in the segment AMB is equal to the given angle C Scholium. If the given angle were a right angle, the required segment would be a semicircle described on AB as a diameter. PROBLEM XVII. Two angles being given, to find their common measure, and by means of it, their ratio in numbers. comparison of two straight lines (B. II., D. 4); since an arc may be cut off from an arc of the same radius, as a straight same as that of the given angles (P. 17); and if DO is the common measure of the arcs, the angle DAO will be that of the angles. Scholium. According to this method, the absolute value of an angle may be found by comparing the arc which measures it, to a quarter circumference. For example, if a quarter circumference is to the angle A as 3 to 1, then, the angle A will be of one right angle, or of four right angles. 12 It may also happen, that the arcs compared have no common measure; in which case, the numerical ratios of the angles will only be found approximatively with more or less correctness, according as the operation is continued. a greater or less number of times. BOOK IV. PROPORTIONS OF FIGURES-MEASUREMENT OF AREAS. DEFINITIONS. 1. SIMILAR FIGURES are those which are mutually equiangular (B. I., D. 22), and have their sides about the equal angles, taken in the same order, proportional. 2. In figures which are mutually equiangular, the angles which are equal, each to each, are called homologous angles: and the sides which are like situated, in respect to the. equal angles, are called homologous sides. 3. AREA, denotes the superficial contents of a figure. The area of a figure is expressed numerically by the number of times which the figure contains some other figurę regarded as a unit of measure. 4. EQUIVALENT FIGURES are those which have equal areas. The term equal, when applied to quantity in general, denotes an equality of measure; but when applied to geometrical figures it denotes an equality in every respect; and such figures when applied the one to the other, coincide in all their parts (A. 14). The term equivalent, denotes an equality in one respect only; viz.: an equality between the measures of figures. The sign, denotes equivalency, and is read, is equivalent to. 5. Two sides of one figure are said to be reciprocally proportional to two sides of another, when one of the sides of the first is to one of the sides of the second, as the remaining side of the second is to the remaining side of the first. |