two chords intersect each other in a circle, the segments are reciprocally proportional. Let the chords AB and CD intersect at 0: then will, Draw AC and BD. In the triangles ACO, DOB, the angles at O are equal, being vertical angles (B. I., P. 4): the angle A is equal to the angle D, because both are inscribed in the same segment (B. III., P. 18, c. 1); for the same reason the angle C=B; B the triangles are therefore similar (P. 18), and the homologous sides give the proportion AO DO :: Cor. Therefore, AOX OBDOX CO: hence, the rectangle of the two segments of one chord is equivalent to the rectangle of the two segments of the other. PROPOSITION XXIX. THEOREM. If from a point without a circle, two secants be drawn terminating in the concave arc, the whole secants will be reciprocally proportional to their external segments. Let the secants OB, OC, be drawn from the point 0: then will OB : OC For, drawing AC, BD, the triangles AOC, BOD have the angle O common; likewise the angle B⇒C (B. III., P. 18, c. 1); these triangles are therefore similar (P. 18), and their homologous sides give the proportion, OB : оо Cor. Hence, the rectangle OBX OA OCXOD. B A Scholium. This proposition, it may be observed, bears a close analogy to the preceding, and differs from it only as the two chords AB, CD, instead of intersecting each other within, cut each other without the circle. The following proposition may be regarded as a particular case of the proposition just demonstrated. PROPOSITION XXX. THEOREM. If from a point without a circle, a tangent and a secant be drawn, the tangent will be a mean proportional between the secant and its external segment. From the point 0, let the tangent OA, and the secant OC be drawn, then will or, ос -2 For, drawing AD and AC, the triangles DAO, CAO, have the angle O common; also, the angle OAD, formed by a tangent and a chord, is measured by half the arc AD (B. III., P. 21); and the angle C has the same measure (B. III., P. 18); hence, the angle OAD=C (A. 1): therefore, the two triangles are similar, and we have the proportion which gives A A If either angle of a triangle is bisected by a line terminating in the opposite side, the rectangle of the sides about the bisected angle, is equivalent to the square of the bisecting line, together with the rectangle contained by the segments of the third side. In the triangle BAC, let AD bisect the angle A; then will AB× ACAD3+BD×DC. Describe a circle through the three points A, B, C (B. III., PROB. 13, s.); prolong AD till it meets the circumference in E, B and draw CE. The triangle BAD is similar to the triangle EAC; for, by hypothesis, the angle BAD=EAC; also, the angle B=E, D E since they are both measured by half the arc AC (B. III., P. 18); hence, these triangles are similar, and the homologous sides give the proportion BA AE :: AD: AC; : hence, BAXAC≈AE×AD; but AE=AD+DE, and multiplying each of these equals by AD, we have AE× AD¬AD2+AD×DE; now (P. 28, c.), ADX DEBD×DC; hence, finally, BAXACAD2+BD× DC. PROPOSITION XXXII. THEOREM. In any triangle, the rectangle contained by two sides is equivalent to the rectangle contained by the diameter of the circumscribed circle, and the perpendicular let fall on the third side. In the triangle BAC, let AD be drawn perpendicular to BC; and let EC be the diameter of the circumscribed circle: then will ABXACAD× CE. For, drawing AE, the triangles DBA, CAE, are right-angled, the one at D, the other at A: also, the angle B=E (B. III., P. 18, c. 1); these triangles are therefore similar, and we have and hence, Cor. If these equal quantities be multiplied by BC, there will result AB×AC×BC=CE×AD×BC; now, ADXBC is double the area of the triangle (P. 6); therefore, the product of the three sides of a triangle is equal to its area multiplied by twice the diameter of the circumscribed circle. The product of three lines is sometimes represented by a solid, for a reason that will be seen hereafter. Its value is easily conceived, by supposing the lines to be reduced to numbers, and then multiplying these numbers together. Scholium. It may also be demonstrated, that the area of a triangle is equal to its perimeter multiplied by half the radius of the inscribed circle. For, the triangles AOB, BOC, AOC, which have a common vertex at 0, have for their common altitude the radius of the inscribed circle; hence, the sum of these triangles will be equal to the B E D A F sum of the bases AB, BC, AC, multiplied by half the radius OD; hence, the area of the triangle ABC is equal to its perimeter multiplied by half the radius of the inscribed circle. In every quadrilateral inscribed in a circle, the rectangle of the two diagonals is equivalent to the sum of the rectangles of the opposite sides. Let ABCD be a quadrilateral inscribed in a circle, and AC, BD, its diagonals: then we shall have ACX BDABX CD+ADX BC. Take the arc CO=AD, and draw BO, meeting the diagonal AC in I The angle ABD=CBI, since the one has for its measure half of the arc AD (B. III., P. 18), and the other, half of CO, equal to AD; the angle ADB =BCI, because they are subtended by A D B the same arc; hence, the triangle ABD is similar to the triangle IBC, and we have the proportion AD CI :: BD: BC; and consequently, ADX BCCI × BD. D B Again, the triangle ABI is similar to the triangle BDC; for the arc AD being equal to CO, if OD be added to each of them, we shall have the arc AO=DC; hence, the angle ABI is equal to DBC; also, the angle BAI to BDC, because they stand on the same arc; hence, the triangles ABI, DBC, are similar, and the homologous sides give the proportion hence, ΑΙ AB : BD ABX CD=AI ×BD. Adding the two results obtained, and observing that we shall have ADX BC+ABX CD-AC×BD. PROBLEMS RELATING TO THE FOURTH BOOK. PROBLEM I. To divide a given straight line into any number of equal parts, or into parts proportional to given lines. First. Let it be proposed to divide the line AB into five equal parts. Through the extremity A, draw the indefinite straight line AG: take AC of any magnitude, and apply it five times upon AG; join the last point of division G, and the extremity B of the given line, by the straight line GB; then through C, draw CI parallel to GB: F A A D -K E -L M B |