PROBLEM XII. The adjacent sides of a parallelogram, with the angle which they contain, being given, to describe the parallelogram Let A and B be the given sides, and C the given angle. point D, make the angle EDF-- DEGF will be the parallelogram required. For, the opposite sides are equal, by construction; hence the figure is a parallelogram (Book I. Prop. XXIX.): and it is formed with the given sides and the given angle. Cor. If the given angle is a right angle, the figure will be a rectangle; if, in addition to this, the sides are equal, it will be a square. PROBLEM XIII. To find the centre of a given circle or arc. Take three points, A, B, C, any where in the circumference, or the arc; draw AB, BC, or suppose them to be drawn; bisect those two lines by the perpendiculars DE, FG: the point O, where these perpendiculars meet, will be the centre sought (Prop. VI. Sch.). Scholium. The same construction serves for making a circum E A ference pass through three given points A, B, C; and also for describing a circumference, in which, a given triangle ABC shall be inscribed. PROBLEM XIV. Through a given point, to draw a tangent to a given circle. If the given point A lies in the circum- D ference, draw the radius CA, and erect AD perpendicular to it: AD will be the tangent required (Prop. IX.). If the point A lies without the circle, join A and the centre, by the straight line CA: bisect CA in O; 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 (Prop. XVIII. Cor. 2.); therefore AB is a perpendicular at the extremity of the radius CB; therefore it is a tangent. B A Scholium. When the point A lies without the circle, there will evidently be always two equal tangents AB, AD, passing through the point A: they are equal, because the right angled triangles CBA, CDA, have the hypothenuse CA common, and the side CB=CD; hence they are equal (Book I. Prop. XVII.); hence AD is equal to AB, and also the angle CAD to CAB. And as there can be but one line bisecting the angle BAC, 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. Let ABC be the given triangle. Bisect the angles A and B, by the lines AO and BO, meeting in the point 0; from the point O, let fall the perpendiculars OD, OE, OF, on the three sides of the triangle: these perpendiculars will all be equal. For, by construc 譯众 tion, we have the angle DAO=OAF, the right angle ADO= AFO; hence the third angle AOD is equal to the third AOF (Book I. Prop. XXV. Cor. 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 (Book I. Prop. VI.); 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; therefore the three perpendiculars OD, 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 evidently be inscribed in the triangle ABC; for the side AB, being perpendicular to the radius at its extremity, is a tangent; and the same thing is true of the sides BC, AC. Scholium. The three lines which bisect the 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 all the angles inscribed in it, shall be equal to the 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 GO perpendicular to AB, through the middle point G; and from the point O, where these perpendiculars meet, as a centre, with a distance OB, describe a circle: the required segment will be AMB. For, since BF is a perpendicular at the extremity of the radius OB, it is a tangent, and the angle ABF is measured by half the arc AKB (Prop. XXI.). Also, the angle AMB, being an inscribed angle, is measured by half the arc AKB: hence we have AMB=ABF=EBD=Č: hence all the angles inscribed in the segment AMB are equal to the given angle C. F* Scholium. If the given angle were a right angle, the required segment would be a semicircle, described on AB as a diameter. PROBLEM XVII. To find the numerical ratio of two given straight lines, these lines being supposed to have a common measure. Let AB and CD be the given lines. From the greater AB cut off a part equal to the less CD, as many times as possible; for example, twice, with the remainder BE. From the line CD, cut off a part equal to the remainder BE, as many times as possible; once, for example, with the remainder DF. From the first remainder BE, cut off a part equal to the second DF, as many times as possible; once, for example, with the remainder BG. From the second remainder DF, cut off a part equal to BG the third, as many times as possible. B G Continue this process, till a remainder occurs, which is contained exactly a certain number of times in the preceding one. Then this last remainder will be the common measure of the proposed lines; and regarding it as unity, we shall easily find the values of the preceding remainders; and at last, those of the two proposed lines, and hence their ratio in numbers. Suppose, for instance, we find GB to be contained exactly twice in FD; BG will be the common measure of the two proposed lines. Put BG=1; we shall have FD=2: but EB contains FD once, plus GB; therefore we have EB=3: CD contains EB once, plus FD; therefore we have CD=5: and, lastly, AB contains CD twice, plus EB; therefore we have AB=13; hence the ratio of the lines is that of 13 to 5. If the line CD were taken for unity, the line AB would be 3; if AB were taken for unity, CD would be. Scholium. The method just explained is the same as that employed in arithmetic to find the common divisor of two numbers: it has no need, therefore, of any other demonstration. How far soever the operation be continued, it is possible that no remainder may ever be found, which shall be contained an exact number of times in the preceding one. When this happens, the two lines have no common measure, and are said to be incommensurable. An instance of this will be seen after wards, in the ratio of the diagonal to the side of the square. In those cases, therefore, the exact ratio in numbers cannot be found; but, by neglecting the last remainder, an approximate ratio will be obtained, more or less correct, according as the operation has been continued a greater or less number of times. PROBLEM XVIII. Two angles being given, to find their common measure, if they have one, and by means of it, their ratio in numbers. ДД D OE With equal radii describe the arcs CD, EF, to serve as measures for the angles: proceed afterwards in the comparison of the arcs CD, EF, as in the last problem, since an arc may be cut off from an arc of the same radius, as a straight line from a straight line. We shall thus arrive at the common measure of the arcs CD, EF, if they have one, and thereby at their ratio in numbers. This ratio will be the same as that of the given angles (Prop. XVII.); and if DO is the common measure of the arcs, 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 the whole circumference. If the arc CD, for example, is to the circumference, as 3 is to 25, the angle A will be of four right angles, or of one right angle. 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 has been continued a greater or less number of times. |