155. Two points are taken in the diameter of a circle at any equal distances from the center; through one of these draw any chord, and join its extremities and the other point. The triangle so formed has the sum of the squares of its sides invariable. 156. If chords drawn from any fixed point in the circumference of a circle, be cut by another chord which is parallel to the tangent at that point, the rectangle contained by each chord, and the part of it intercepted between the given point and the given chord, is constant. 157. If AB be a chord of a circle inclined by half a right angle to the tangent at A, and AC, AD be any two chords equally inclined to AB, ACAD2 = 2. A B2. 158. A chord POQ cuts the diameter of a circle in Q, in an angle equal to half a right angle; PO2 + O Q2 = 2 (rad. ). 159. Let ACDB be a semicircle whose diameter is AB; and AD, BC any two chords intersecting in P; prove that ABDA. AP+ CB.BP. 160. If ABDC be any parallelogram, and if a circle be described passing through the point A, and cutting the sides AB, AC, and the diagonal AD, in the points F, G, H respectively, shew that AB.AF+AC.AG=AD.AH. 161. Produce a given straight line, so that the rectangle under the given line, and the whole line produced, may equal the square of the part produced. 162. If A be a point within a circle, BC the diameter, and through A, AD be drawn perpendicular to the diameter, and BAE meeting the circumference in E, then BA.BE-BC.BD. 163. The diameter ACD of a circle, whose center is C, is produced to P, determine a point F in the line AP such that the rectangle PF.PC may be equal to the rectangle PD. PA. 164. To produce a given straight line, so that the rectangle contained by the whole line thus produced, and the part of it produced, shall be equal to a given square. 165. Two straight lines stand at right angles to each other, one of which passes through the center of a given circle, and from any point. in the other, tangents are drawn to the circle. Prove that the chord joining the points of contact cuts the first line in the same point, whatever be the point in the second from which the tangents are drawn. 166. A, B, C, D, are four points in order in a straight line, find a point E between B and C, such that AE.EB=ED.EC, by a geometrical construction. 167. If any two circles touch each other in the point O, and lines be drawn through O at right angles to each other, the one line cutting the circles in P, P, the other in Q, Q'; and if the line joining the centers of the circles cut them in A, A'; then BOOK IV. DEFINITIONS. I. A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angular points of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each. II. In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. III. A rectilineal figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle. IV. A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle. V. In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure. VI. A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described. VII. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle. PROPOSITION I. PROBLEM. In a given circle to place a straight line, equal to a given straight line which is not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. It is required to place in the circle ABC a straight line equal to D. Draw BC the diameter of the circle ABC. Then, if BC is equal to D, the thing required is done; for in the circle ABC a straight line BC is placed equal to D. But, if it is not, BC is greater than D; (hyp.) make CE equal to D, (1. 3.) and from the center C, at the distance CE, describe the circle AEF, and join CA. Then CA shall be equal to D. Because C is the center of the circle AEF, therefore D is equal to CA. (ax. 1.) Wherefore in the circle ABC, a straight line CA is placed equal to the given straight line D, which is not greater than the diameter of the circle. Q.E.F. PROPOSITION II. PROBLEM. In a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle. It is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. Draw the straight line GAH touching the circle in the point A, (III. 17.) and at the point A, in the straight line ÀH, make the angle HAC equal to the angle DEF; (1. 23.) and AC is drawn from the point of contact, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle: (III. 32.) but HAC is equal to the angle DEF; (constr.) therefore also the angle ABC is equal to DEF: (ax. 1.) for the same reason, the angle ACB is equal to the angle DFE: therefore the remaining angle BAC is equal to the remaining angle EDF: (1. 32. and ax. 3.) wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. PROPOSITION III. PROBLEM. Q. E. F. About a given circle to describe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle. It is required to describe a triangle about the circle ABC equiangular to the triangle DEF. Produce EF both ways to the points G, H; at the point K in the straight line KB, and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC. (III. 17.) Then LMN shall be the triangle required. Because LM, MN, NL touch the circle ABC in the points A,B, C, to which from the center are drawn KA, KB, KC, therefore the angles at the points A, B, C are right angles: (III. 18.) and because the four angles of the quadrilateral figure AMBK are equal to four right angles, for it can be divided into two triangles; and that two of them KAM, KBM are right angles, therefore the other two AKB, AMB are equal to twò right angles: (ax. 3.) but the angles DEG, DEF are likewise equal to two right angles; (1. 13.) therefore the angles AKB, AMB are equal to the angles DEG, DEF; (ax. 1.) of which AKB is equal to DEG; (constr.) wherefore the remaining angle AMB is equal to the remaining angle DEF. (ax. 3.) In like manner, the angle LNM may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal to the remaining angle EDF: (1.32 and ax. 3.) therefore the triangle LMN is equiangular to the triangle DEF: and it is described about the circle ABC. Q. E. F. Bisect the angles ABC, BCA by the straight lines BD, CD meeting one another in the point D, (I. 9.) from which draw DE, DF, DG perpendiculars to AB, BC, CA. (1.12.) And because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, and that the right angle BED is equal to the right angle BFD; (ax.11.) therefore the two triangles EBD, FBD have two angles of the one equal to two angles of the other, each to each; and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore their other sides are equal; (I. 26.) for the same reason, DG is equal to DF: therefore the three straight lines DE, DF, DG are equal to one another; and the circle described from the center D, at the distance of any of them, will pass through the extremities of the other two, and touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches the circle: (III. 16.) therefore the straight lines AB, BC, CA do each of them touch the circle, and therefore the circle EFG is inscribed in the triangle ABC. Q.E.F. |