of the line ED must be equally distant from A and B ; hence the point D, common to these two lines, must be at equal distances from the three points A, B, and C, and the center of a circle passing through them. Scholium. There is but one such circle. For its center could not be out of the line FA (th. 18), nor out of EC, and they intersect in but one point D.* Note. The problem is the same, in effect, when it is required To describe the circumference of a circle through three given points A, B, C, or to find the center of a given circle or arc. Draw chords BA, BC, and bisect these chords perpendicularly by lines meeting in D, which will be the center. (See last diagram.) PROBLEM XVIII. An isosceles triangle ABC being given, to describe another on the same base AB, whose vertical angle shall be only half the vertical angle C. From C as a center, with the distance CA, describe the circle ABE. Bisect AB in D, join DC, and produce to the circumference at E; join EA and EB, and ABE shall be the isosceles triangle required. E A D B For every point of the perpendicular DE is equally distant from A and B (th. 17, corol. 1); hence the side EA must be equal to the side EB of the triangle AEB, which is, therefore, isosceles, and the angle ACB at the center must be double of the angle AEB at the circumference, for they both stand on the same segment AB. *If the given triangle be acute angled, the center of the circle will be within it; and if the triangle be equilateral, as in the diagram, the center of the circle will be the center of the triangle, and the perpendiculars at the middle of the sides will pass through the vertices of the opposite angles. If the triangle be obtuse angled, the center of the circle will fall without; if right angled, the center will fall upon the hypothenuse. PROBLEM XIX. Given an isosceles triangle AEB, to erect another on the same base AB, which shall have double the vertical angle E. Describe a circle about the triangle AEB, find its center C, and join CA, CB, and ACB is the triangle required. The angle C at the center is double of the angle E at the circumference, and the triangle ACB is isosceles; for the sides CA, CB, being radii of the A same circle, are equal. PROBLEM XX. E D E To draw a tangent to a circle, through a given point A. 1. When the given point A is in the E A circumference of the circle, join A and the center O; perpendicular to which draw BAC, and it will be the tangent, by th. 36. BA E G D C 2. When the given point A is out of the circle, draw AO to the center O; on which, as a diameter, describe a semicircle, cutting the given circumference in D; through which draw BADC, which will be the tangent, as required. For join DO. Then the angle ADO, in a semicircle, is a right angle, and, consequently, AD is perpendicular to the radius DO, or is a tangent to the circle (th. 36). Scholium. The circle ADO cuts the given circle in two points; and there will be two tangents, AD and AE, to the given circle from the same point A, without. These tangents are equal in length, and the line joining the point without and the center bisects the angle which the tangents make with each other; for the right-angled triangles ADO, AEO, having the = side OA common, and the side OD PROBLEM XXI. OE being radii equal .. AD = On a given line AB to describe a segment of a circle capable of containing a given angle. At the ends of the given line make angles DAB, DBA, each equal to the given angle C. Then draw AE, BE perpendicular to AD, BD; and with the A center E, and radius EA or EB, describe a circle; so shall AFB be the segment required, as any angle F made in it will be equal to the given angle C. D F B For the two lines AD, BD, being perpendicular to the radii EA, EB (by construction), are tangents to the circle (th. 36); and the angle A or B, which is equal to the given angle C by construction, is equal to the angle F, being all three measured by half the arc AB (th. 38 and 39).* Scholium. One of the lines AD, BD may be omitted, and a perpendicular drawn at the middle of AB to meet the other at the point E. This problem is particularly useful in the survey of harbors. Three points on the shore are chosen, which, being connected by lines, form a triangle; then from a boat, where a sounding is to be made, the angles subtended by two of the sides of this triangle are measured with a sextant. To transfer this to a map, there must first be made upon the paper the triangle whose sides unite the three points upon the shore. Then upon one of the sides of this triangle, by the above problem, make a segment capable of containing one of the observed angles, and upon the other a segment capable of containing the other observed angle; the point in which the arcs of these two segments intersect will be the point on the map corresponding to that where the sounding was made, and there the depth in fathoms or feet may be written down. PROBLEM XXII. To inscribe an equilateral triangle in a given circle. Through the center C draw any diameter AB. From the point B as a center, with the radius BC of the given circle, describe an arc DCE. Join AD, AE, DE, and ADE is the equilateral triangle sought. A B Join DB, DC, EB, EC. Then DCB is an equilateral triangle, having each side equal to the radius of the given circle. In like manner, BCE is an equilateral triangle. But the angle ADE is equal to the angle ABE or CBE, standing on the same arc AE; also, the angle AED is equal to the angle CBD, on the same arc AD; hence the triangle DÃE has two of its angles, ADE, AED, equal to the angles of an equilateral triangle, and therefore the third angle at A is also equal to the same; so that the triangle is equiangular, and therefore equilateral. PROBLEM XXIII. To inscribe a circle in a given triangle ABC. Bisect any two angles C and B with the two lines CD, BD. From the intersection D, which will be the center of the circle, draw the perpendiculars DE, DF, DG, and they will be the radii of the circle required. B A G E F For, since the sides CB, CA, are to be tangents, the line CD, bisecting the angle which they form, must pass through the center. (Prob. 20, schol.) For a similar reason, BD must pass through the center. Hence it is at the intersection D of these two lines. PROBLEM XXIV. To inscribe a square in a given circle. Draw two diameters AC, BD, crossing at right angles in the center E. Then join the four extremities A, B, C, D with right lines, and these will form the inscribed square ABCD. For the four right-angled triangles AEB, BEC, CED, DEA are A B E identical, because they have the sides EA, EB, EC, ED all equal, being radii of the circle, and the four included angles at E all equal, being right angles, by the construction. Therefore, all their third sides, AB, BC, CD, DA, are equal to one another, and the figure ABCD is equilateral. Also, all its four angles, A, B, C, D, are right ones, being angles in a semicircle. Consequently, the figure is a square. PROBLEM XXV. To find a fourth proportional to three given lines, AB, AC, AD. Place two of the given lines AB, A AC, or their equals, to make any an- A gle at A; and on AB set off, or place, A the other line AD, or its equal. Join BC, and parallel to it draw DE; so shall AE be the fourth proportional, as required. E B C D D B For, because of the parallels BC, DE, the two sides AB, AC are cut proportionally (th. 61); so that AB: AC:: AD: AE. PROBLEM XXVI. To find a mean proportional between two lines AB, BC. |