scribed about the triangle; then will ABX AC be equivalent to ADX CE. For, joining AE, the angle EAC is a right angle, being inscribed in a semicircle (Prop. XVIII. Cor. 2, Bk. III.); and the angles B and E D A E are equal, being measured by half of the same arc, A C (Prop. XVIII. Cor. 1, Bk. III.); hence the two rightangled triangles are similar (Prop. XXII. Cor.), and give the proportion AB: CE::AD: AC; hence ABX AC=CEX AD. 290. Cor. If these equals be multiplied by BC, we shall have ABX ACX BC= CEX ADX BC. But ADX BC is double the area of the triangle (Prop. VI.); therefore the product of the three sides of a triangle is equal to its area multiplied by twice the diameter of the circumscribed circle. PROPOSITION XXXVIII.-THEOREM. 291. The rectangle contained by the diagonals of a quadrilateral inscribed in a circle is equivalent to the sum of the two rectangles of the opposite sides. Let ABCD be any quadrilateral inscribed in a circle, and A C, BD its diagonals; then the rectangle A CX BD is equivalent to the sum of the two rectangles ABX CD, AD X BC. For, draw BE, making the angle ABE equal to the angle CBD; B C A E D to each of these equals add the angle EBD, and we shall have the angle ABD equal to the angle EBC; and the angle ADB is equal to the angle BCE, being in the same segment B (Prop. XVIII. Cor. 1, Bk. III.); AD: BD::CE: BC; C E Again, since the angle ABE is equal to the angle CBD, and the angle B A E is equal to the angle B D C, being in the same segment (Prop. XVIII. Cor. 1, Bk. III.), the triangles A BE, BCD are similar; hence, AB: AE:: BD: CD; and consequently, ABX CD AEX BD. = By adding the corresponding terms of the two equations obtained, and observing that BDX AE+BD × CE=BD (AE÷CE)=BD × AC, we have BDX AC ABX CD+AD × BC. = PROPOSITION XXXIX.-THEOREM. 292. The diagonal of a square is incommensurable with its side. Let ABCD be any square, and A C its diagonal; then AC is incommensurable with the side A B. D E To find a common measure, if there be one, we must apply A B, or its equal CB, to CA, as often as it can be done. In order to do this, from the point C as a centre, with a radius C B, describe the semicircle F B E, and produce AC to E. It is evident that CB is contained once in AC, A G B with a remainder A F, which remainder must be compared with BC, or its equal, A B. The angle ABC being a right angle, A B is a tangent to the circumference, and A E is a secant drawn from the same point, so that (Prop. XXXV.) AF:AB::AB: AE. Hence, in comparing AF with A B, the equal ratio of AB to A E may be substituted; but A B or its equal CF is contained twice in A E, with a remainder A F; which remainder must again be compared with A B. Thus, the operation again consists in comparing A F with A B, and may be reduced in the same manner to the comparison of A B, or its equal CF, with AE; which will result, as before, in leaving a remainder A F; hence, it is evident that the process will never terminate; consequently the diagonal of a square is incommensurable with its side. 293. Scholium. The impossibility of finding numbers to express the exact ratio of the diagonal to the side of a square has now been proved; but, by means of the continued fraction which is equal to that ratio, an approximation may be made to it, sufficiently near for every practical purpose. BOOK V. PROBLEMS RELATING TO THE PRECEDING BOOKS. PROBLEM I. 294. To bisect a given straight line, or to divide it into two equal parts. Let A B be a straight line, which it is required to bisect. E A B XD From the point A as a centre, with a radius greater than the half of A B, describe an arc of a circle; and from the point B as a centre, with the same radius, describe another arc, cutting the former in the points C and D. Through C and D draw the straight line CD; it will bisect A B in the point E. For the two points C and D, being each equally distant from the extremities A and B, must both lie in the perpendicular raised from the middle point of AB (Prop. XV. Cor., Bk. I.). Therefore the line CD must divide the line A B into two equal parts at the point E. PROBLEM II. 295. From a given point, without a straight line, to draw a perpendicular to that line. Let A B be the straight line, and let C be a given point without the line. From the point C as a centre, and with a radius sufficiently great, describe an arc cutting the line AB in two points, A and B; then, from the points A and B as centres, with a radius greater than half of A B, describe two arcs cutting each other in D, and draw the straight line CD; it will be the perpendicular required. For, the two points C and D are each equally distant from the points A and B; hence, the line CD is a perpendicular passing through the middle of A B (Prop. XV. Cor., Bk. I.). PROBLEM III. 296. At a given point in a straight line to erect a perpendicular to that line. Let A B be the straight line, and let D be a given point in it. C A Ꭰ B In the straight line AB, take the points A and B at equal distances from D; then from the points A and B as centres, with a radius greater than AD, describe two arcs cutting each other at C; through C and D draw the straight line CD; it will be the perpendicular required. For the point C, being equally distant from A and B, must be in a line perpendicular to the middle of A B (Prop. XV. Cor., Bk. I.); hence CD has been drawn perpendicular to A B at the point D. 297. Scholium. The same construction serves for making a right angle, A D C, at a given point, D, on a given straight line, A B. |