If we had taken BP AD, and had drawn CKP, we should have found by similar reasoning CP X CAAB × 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 BO × BD and CP X CA, are to each other as BD is to CA; therefore BD: CA:: AB × BC+AD × DC: AB × AD + BC × CD; 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 Fig. 137. Fig. 138. 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. 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 AI is a fifth part of AB. 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 K into parts proportional to the given lines P, Q, R. For, on account of the parallels CI, DK, EB, the parts AI, 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 .4, B; for it would be the same as the fourth proportional to the three lines A, B, B. PROBLEM. 239. To find a mean proportional between two given lines A and B (fig. 140). Solution. On the indefinite line DF take DEA, and EFB; on the whole line DF, as a diameter, describe the semicircumference DGF; at the point E erect upon the diameter the perpendicular EG meeting the circumference in G; EG will be the mean proportional sought. For the perpendicular GE, let fall from the point in the circumference upon the diameter, is a mean proportional between the two segments of the diameter DE, EF (215), and these two segments are equal to the two given lines A and B. PROBLEM. Fig. 140. 240. To divide a given line AB (fig. 141) into two parts in such Fig. 141 a manner, that the greater shall be a mean proportional between the whole line and the other part. Solution. At the extremity B of the line AB erect the perpendicular BC equal to half of AB; from the point C as a centre, and with the radius CB, describe a circle; draw AC cutting the circumference in D, and take AF = AD; the line AB will be divided at the point F in the manner required; that is, AB: AF:: AF: FB. For AB, being a perpendicular to the radius CB at its extremity CB, is a tangent; and, if AC be produced till it meet the circumference in E. we shall have ᎯᎬ : ᏁᏴ :: ᏁᏴ : ᎯᎠ (228), and hence AE-AB: AB:: AB-AD: AD (IV): But, since the radius BC is half of AB, the diameter DE is equal to AB, and consequently AE AB = AD = AF; also, since AFAD, AB-ADFB; therefore, and by inversion AF: AB:: FB: AD or AF, AB: AF :: AF : FB. 241. Scholium. When a line is divided in this manner, it is said to be divided in extreme and mean ratio. Its application will be seen hereafter. It may be remarked, that the secant AE is divided in extreme and mean ratio at the point D; for since AB = DE, AE: DE:: DE: AD. Fig. 142. Fig. 143. Fig. 144. Fig. 145. PROBLEM. 242. Through a given point A (fig. 142) in a given angle BCD, to draw a line BD in such a manner that the parts AB, AD, comprehended between the point A and the two sides of the angle shall be equal. Solution. Through the point A draw AE parallel to CD, take BE CE, and through the points B and A draw BAD, which will be the line required. = 243. To make a square equivalent to a given parallelogram, or to a given triangle. Solution. 1. Let ABCD (fig. 143) be the given parallelogram, AB its base, and DE its altitude; between AB and DE find a mean proportional XY (239); the square described upon XY will be equivalent to the parallelogram ABCD. For, by construction, AB: XY:: XY: DE; hence --2 XY=AB × DE ; but AB × DE is the measure of the parallelogram, and XY is that of the square, therefore they are equivalent. 2. Let ABC (fig. 144) be the given triangle, BC its base, and AD its altitude; find a mean proportional between BC and half of AD, and let XY be this mean proportional; the square described upon XY will be equivalent to the triangle ABC. For, since BC: XY:: XY : ¦ AD, XY2= BC × AD; therefore the square described upon XY is equivalent to the triangle ABC. PROBLEM 244. Upon a given line AD (fig. 145) to construct a rectangle ADEX equivalent to a given rectangle ABFC. Solution. Find a fourth proportional to the three lines AD, AB, AC (137), and let AX be this fourth proportional; the rectangle contained by AD and AX will be equivalent to the rectangle ABFC. = For, since AD: AB::AC: AX, AD × AX ABX AC; therefore the rectangle ADEX is equivalent to the rectangle ABFC. PROBLEM. 245. To find in lines the ratio of the rectangle of two given lines A and B (fig. 148) to the rectangle of two given lines C and D. Solution. Let X be a fourth proportional to the three given lines B, C, D; the ratio of the two lines A and X will be equal to that of the two rectangles Ax B, C × D. For, since B: C:: D: X, CX D=BX X; therefore 246. Corollary. In order to obtain the ratio of the squares described upon two lines A and C, find a third proportional X to the lines A and C, so that we may have the proportion Fig. 148. PROBLEM. 247. To find in lines the ratio of the product of three given lines A, B, C (fig. 149), to the product of three given lines P, Q, R. Solution. Find a fourth proportional X to the three given lines P, A, B ; and a fourth proportional Y to the three given lines C, Q, R. The two lines X and Y will be to each other as the products A×B× C, PXQ × R. For, since P: A: : B : X, A × B =P × X; and, by multiplying each of these by C, we shall have AxBX C=CxPx X. In like manner, since C: QR: Y, Q× R = C × Y; and, by multiplying each of these by P, we shall have PXQXR=PxCxY; therefore the product AxBX C: the product Px Qx R:: CxPxX: Px CxY:: X: Y. Fig. 149. PROBLEM. 248. To make a triangle equivalent to a given polygon. Solution. Let ABCDE (fig. 146) be the given polygon. Fig. 146. Draw the diagonal CE, which cuts off the triangle CDE; through the point D draw DF parallel to CE to meet AE pro Geom. 10 |