Let the chords AB and CD intersect at O: then will AO : DO : : OC : OB. Draw AC and BD. In the triangles ACO, BOD, the angles at O are equal, being vertical; the angle A is equal to the angle D, because both are inscribed in the same segment (Book III. Prop. XVIII. Cor. 1.); for the same reason the angle CB; the triangles are therefore similar, and the homologous sides give the proportion AO: DO: CO: OB. A B Cor. Therefore AO.OB-DO.CO: hence the rectangle ander the two segments of the one chord is equal to the rectangle under the two segments of the other. PROPOSITION XXIX. THEOREM. If from the same 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 :: OD: OA. For, drawing AC, BD, the triangles OAC, OBD have the angle O common; likewise the angle B C (Book III. Prop. XVIII. Cor. 1.); these triangles are therefore similar; and their homologous sides give the proportion, OB OC :: OD : OA. Cor. Hence the rectangle OA.OB is equal to the rectangle OC.OD. B D Scholium. This proposition, it may be observed, bears a great 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 also be regarded as a particular case of the proposition just demonstrated. PROPOSITION XXX. THEOREM. If from the same 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 O, let the tangent OA, and the secant OC be be drawn; then will OC: OA :: OA: OD, or OA2=OC.OD. For, drawing AD and AC, the triangles Q OAD, OAC, have the angle O common; also the angle OAD, formed by a tangent and a chord, has for its measure half of the arc AD (Book III. Prop. XXI.); and the angle Chas the same measure: hence the angle OAD A C; therefore the two triangles are similar, and we have the proportion OC : OA :: AO OD, which gives OA2=OC.OD. C PROPOSITION XXXI. THEOREM. If either angle of a triangle be bisected by a line terminating in the opposite side, the rectangle of the sides including 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.AC=AD2+BD.DC. Describe a circle through the three points A, B, C ; produce AD till it meets the circumference, and draw CE. The triangle BAD is similar to the trian- B gle EAC; for, by hypothesis, the angle BAD EAC; also the angle B=E, since they are both measured by half of the arc AC; hence these triangles are similar, and D the homologous sides give the proportion BA : AE : : AD : AC; hence BA.AC-AE.AD; but AE=AD+DE, and multiplying each of these equals by AD, we have AE.AD=AD2+ AD.DE; now AD.DE=BD.DC (Prop. XXVIII.); hence, finally, BA.AC AD2+BD.DC. PROPOSITION XXXII. THEOREM. In every 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 upon the third side. In the triangle ABC, let AD be drawn perpendicular to BC; and let EC be the diameter of the circumscribed circle; then will AB.AC=AD.CE. For, drawing AE, the triangles ABD, AEC, are right angled, the one at D, the other at A: also the angle B-E; these triangles are therefore similar, and they give the proportion AB : CE :: AD: AC; and hence AB.AC-CE.AD. Cor. If these equal quantities be multiplied by the same quantity BC, there will result AB.AC.BC=CE.AD.BC; now AD.BC is double of the area of the triangle (Prop. VI.); therefore the product of 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 called a solid, for a reason that shall be seen afterwards. Its value is easily conceived, by imagining that the lines are reduced into numbers, and 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 O, have for their common altitude the radius of the inscribed circle; hence the sum of these triangles will be equal to the sum of the bases AB, BC, AC, multiplied by half the radius D B F E OD; hence the area of the triangle ABC is equal to the perimeter multiplied by half the radius of the inscribed circle PROPOSITION XXXIII. THEOREM. 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. In the quadrilateral ABCD, we shall have AC.BD=AB.CD+AD.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, and the other, half of CO, equal to AD; the angle ADB=BCI, because they are A both inscribed in the same segment AOB; hence the triangle ABD is similar to the triangle IBC, and we have the : proportion AD CI: BD BC; hence AD.BC=CI.BD 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 are inscribed in the same segment; hence the triangles ABI, DBC, are similar, and the homologous sides give the proportion AB : BD :: AI: CD; hence AB.CD=AI.BD. Adding the two results obtained, and observing that AI.BD+CI.BD=(AI+CI).BD=AC.BD, we shall have AD.BC+AB.CD=AC.BD. I 13 PROBLEMS RELATING TO THE FOURTH BOOK. PROBLEM 1. 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; and taking AC of any magnitude, apply it five times upon AG; join the last point of division G, and the extremity B, by the straight line GB; then draw CI parallel to GB: AI will be the fifth part of the line AB; and thus, by applying AI five times upon AB, the line AB will be divided into five equal parts. For, since CI is parallel to GB, the sides AG, AB, are cut proportionally in C and I (Prop. XV.). But AC is the fifth part of AG, hence AI is the fifth part of AB, P Secondly. Let it be proposed to divide the line AB into parts proportional to the given lines P, Q, R. Through A, draw the indefinite line AG; make AC=. Q r, CD=Q, DE=R; join R the extremities E and B ; and through the points C, D, draw CI, DF, parallel to EB; the line AB will be divided into parts AI, IF, FB, proportional to the given lines P, Q, R. For, by reason of the parals CI, DF, EB, the parts AI, IF, FB, are proportional to the parts AC, CD, DE; and by Construction, these are equal to the given lines P, Q, R. |