secants will be reciprocally proportional to the parts without the Demonstration. Join AC and BD. The triangles OAC, OBD, have the angle O common; moreover the angle B = C (126); therefore the triangles are similar; and the homologous sides give the proportion OB: OC:: OD : OA. 226. Corollary. The rectangle OA × OB = OC × OD. 227. Scholium. It may be remarked, that this proposition has great analogy with the preceding; the only difference is, that the two chords AB, CD, instead of intersecting each other in the circle, meet without it. The following proposition may also be regarded as a particular case of this. Fig. 132. THEOREM. 228. If, from the same point O (fig. 132), taken without the circle, a tangent OA be drawn, and a secant OC, the tangent will be a mean proportional between the secant and the part without 2 the circle; that is, OC : OA :: 'OA : OD, or, OA = OC × OD. Demonstration. By joining AD and AC, the triangles OAD, OAC, have the angle O common; moreover, the angle OAD formed by a tangent and a chord (131) has for its measure the half of the arc AD, and the angle C has the same measure; consequently the angle OAD = C; therefore the two triangles are similar, and OC:OA :: OA:OD, which gives OA = OC× OD. 2 Fig. 133. THEOREM. 229. In any triangle ABC (fig. 133), if the angle A be bisected by the line AD, the rectangle of the sides AB, AC, will be equal to the rectangle of the segments BD, DC, plus the square of the bisecting line AD. Demonstration. Describe a circle, the circumference of which shall pass through the points A, B, C; produce AD till it meet the circumference, and join CE. The triangle BAD is similar to the triangle EAC; for, by hypothesis, the angle BAD = EAC; moreover the angle B =E, since they have each for their measure the half of the arc AC; consequently the triangles are similar; and the homologous sides give the proportion ᏴA : ᏁᎬ :: ᎯᎠ : ᏁᏟ ; whence BA × AC=AE × AD; but AE = AD + DE, and, by 2 multiplying each by AD, we have AE × AD=AD+AD × DE; besides, AD × DE = BD × DC (224); therefore 230. In every triangle ABC (fig. 134) the rectangle of two of Fig. 134. the sides AB, AC, is equal to the rectangle contained by the diameter CE of the circumscribed circle and the perpendicular AD, let fall upon the third side BC. Demonstration. Join AE, and the triangles ABD, AEC, are right-angled, the one at D, and the other at A; moreover the angle B = E; consequently the triangles are similar; and they give the proportion, AB : CE :: AD:AC; whence ᏁᏴ × ᏁᏟ = CE × AD. 231. Corollary. If these equal quantities be multiplied by BC, we shall have AB × AC × BC = CE × AD × BC. Now AD × BC is double the surface of the triangle (176); therefore the product of the three sides of a triangle is equal to the surface multiplied by double the diameter of the circumscribed circle. The product of three lines is sometimes called a solid, for a reason that will be given hereafter. The value of this product is easily conceived by supposing the three lines reduced to numbers, and these numbers multiplied together. 232. Scholium. It may be demonstrated, also, that the surface of a triangle is equal to its perimeter multiplied by half of the radius of the inscribed circle. For the triangles AOB, BOC, AOC (fig. 87), which have Fig. 87. their common vertex in O, have for their common altitude the radius of the inscribed circle; consequently the sum of these triangles will be equal to the sum of the bases AB, BC, AC, multiplied by half of the radius OD; therefore the surface of the triangle ABC is equal to the product of its perimeter by half of the radius of the inscribed circle. Fig 135. THEOREM. 233. In every inscribed quadrilateral figure ABCD (fig. 135), the rectangle of the two diagonals AC, BD, is equal to the sum of the rectangles of the opposite sides; that is, AC × BD = AB × CD + AD × BC. Demonstration. Take the arc CO=AD, and draw BO meeting the diagonal AC in I. The angle ABD = CBI, since 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 inscribed in the same segment AOB; consequently the triangle ABD is similar to the triangle IBC, and AD : CI :: BD : BC; whence AD × BC = CI × BD. Again, the triangle ABI is similar to the triangle BDC; for, the arc AD being equal to CO, if we add to each of these OD, we shall have the arc AO=DC; consequently the angle ABI is equal to DBC; moreover the angle BAI= BDC, because they are inscribed in the same segment; therefore the triangles ABI, BDC, are similar, and the homologous sides give the proportion AB: BD :: AI : CD; whence, AB × CD = AI × BD. Adding the two results above found, and observing that AI × BD+ CI × BD = (AI+ CI) × BD=AC × BD, we have AD × BC + AB × CD=AC × BD. 234. Scholium. We may demonstrate, in a similar manner, another theorem with respect to an inscribed quadrilateral figure. The triangle ABD being similar to BIC, BD : BC :: AB : BI, whence BI × BD = BC × AB. If we join CO, the triangle ICO, similar to ABI, is similar to BDC, and gives the proportion BD: CO :: DC : OI, whence we have OI X BD = CO × DC, or, CO being equal to AD, OI × BD = AD × DC. Adding these two results, and observing that BI × BD+OI × BD reduces itself to BOX BD, we obtain BO × BD = AB × BC + AD × DC. If we had taken BP = AD, and had drawn CKP, we should have found by similar reasoning CP × CA=AB × AD + BC × CD. But the arc BP being equal to CO, if we add to each BC, we shall have the arc CBP = BCO; consequently the chord CP is equal to the chord BO, and the rectangles BOX BD and CP × CA, are to each other as BD is to CA; therefore BD : CA : : ᏁᏴ × BC + ᎯᎠ × ᎠᏟ : ᏁᏴ X ᎯᎠ + BC × ᏟᎠ; that is, the two diagonals of an inscribed quadrilateral figure are to each other as the sums of the rectangles of the sides adjacent to their extremities. By means of these two theorems the diagonals may be found, when the sides are known. THEOREM. 235. Let P (fig. 136) be a given point within a circle in the Fig. 136. radius AC, and let there be taken a point Q without the circle in the same radius produced such that CP: CA::CA:CQ; if, from any point M of the circumference, straight lines MP, MQ, be drawn to the points P and Q, these straight lines will always be in the same ratio, and we shall have MP:MQ::AP : AQ. Demonstration. By hypothesis CP : CA :: CA: CQ; putting CM in the place of CA, we shall have CP: CM :: CM :CQ; consequently the triangles CPM, CQM, having an angle of the one equal to an angle of the other, and the sides about the equal angles proportional, are similar (208); therefore MP : MQ :: CP : CM or CA. Problems relating to the Third Section. PROBLEM. 236. To divide a given straight line into any number of equal parts, or into parts proportional to any given lines. Fig. 137. Solution. 1. Let it be proposed to divide the line AB (fig. 137) into five equal parts; through the extremity A draw the indefinite straight line AG, and take AC of any magnitude whatever, and apply it five times upon AG; through the last point of the division G draw GB, and through C draw CI parallel to GB; Al will be a fifth part of the line AB, and, 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 (196). But AC is a fifth part of AG, therefore Al is a fifth part of AB. Fig. 138. 2. Let it be proposed to divide the line AB (fig. 138) into parts proportional to the given lines P, Q, R. Through the extremity A draw the indefinite straight line AG, and take AC = P, CD = Q, DE = R ; join EB, and through the points C, D, draw CI, DK, parallel to EB; the line AB will be divided at I and Kinto parts proportional to the given lines P, Q, R. For, on account of the parallels CI, DK, EB, the parts Al1, IK, KB, are proportional to the parts AC, CD, DE (196); and, by construction, these are equal to the given lines P, Q, R. PROBLEM. 237. To find a fourth proportional to three given lines A, B, C Fig. 139. (fig. 139). Solution. Draw the two indefinite lines DE, DF, making any angle with each other. On DE take DA=A, DB = B; and upon DF take DC = C; join AC, and through the point B draw BX parallel to AC; DX will be the fourth proportional required. For, since BX is parallel to AC, DA : DB :: DC : DX; but the three first terms of this proportion are equal to the three given lines; therefore DX is the fourth proportional required. 238. Corollary. We might find, in the same manner, a third proportional to two given lines A, B; for it would be the same as the fourth proportional to the three lines A, B, C. |