now, ADX BC 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 D B 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 BD ABX 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 = 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 BC CIX 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 stand on the same arc; hence, the triangles ABI, DBC, are similar, and the homologous sides give the proportion hence, AB BD :: AI: CD; : ABX CDAI ×BD. Adding the two results obtained, and observing that we shall have ADX BC+ABX CDACX 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: E A D -K -L -M B AI will be the fifth part of the line AB; and by apply. ing 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 (P. 15). But AC is the fifth part of AG, hence, AI is the fifth part of AB. 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 A0=P, CD=Q, DE=R; join the extremities E and B; and through the points C and 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 parallels CI, DF, EB, the parts AI, IF, FB, are proportional to the parts AC, CD, DE (p. 15, c. 2); and by construction, these are equal to the given lines P, Q, R. PROBLEM II. To find a fourth proportional to three given lines, A, B, C. the point B, draw BX parallel to AC; and DX will be the fourth proportional required. For, since BX is parallel to AC, we have the proportion (P. 15, c. 1), DC: ᎠᎪ : ᎠᏴ :: ᎠᎴ : DX; now, the first three terms of this proportion are equal to the three given lines: consequently, DX is the fourth proportional required. Cor. A third proportional to two given lines, A, B, may be found in the same manner, for it will be the same as a fourth proportional to the three lines, A, B, B. PROBLEM III. To find a mean proportional between two given lines A and B, Upon the indefinite line DF, take DE=A, and EF-B; and upon the whole line DF, as a diameter, describe the semicircumference DGF; at the point E, erect, upon the diameter, the per D E B AH pendicular EG meeting the semicircumference in G; EG will be the mean proportional required. For, the perpendicular EG, let fall from a point in the circumference upon the diameter, is a mean proportional between the two segments of the diameter DE, EF (P. 23, c.); and these segments are equal to the given lines A and B. PROBLEM IV. To divide a given line into two such parts, that the greater part shall be a mean proportional between the whole line and the other part. Let AB be the given line. At the extremity B, erect the perpendicular BC, equal to the half of AB; from the point C, as a centre, with the radius CB, describe a semicircle; draw AC cutting the circumference in D; A F B and take AF-AD: then F will be the point of division, and we shall have, AB: AF :: AF : FB. For, AB being perpendicular to the radius at its extremity, is a tangent (B. III., P. 9); and if AC be prolonged till it again meets the circumference, in E, we shall have But, since the radius is the half of AB, the diameter DE is equal to AB, and consequently, AE-AB-AD=AF; also, because AF-AD, we have AB-AD=FB: hence, AF : AB :: FB: AD, or AF; whence, by inversion, AB : AF :: AF : FB. Scholium. This sort of division of the line AB, viz., so that the whole line shall be to the greater part as the greater part is to the less, is called division in extreme and mean ratio. It may further be observed, that the secant AE is divided in extreme and mean ratio at the point D; for, since AR= DE, we have, AE : DE :: DE : AD. PROBLEM V. Through a given point, in a given angle, to draw a line so that the segments comprehended between the point and the two sides of the angle, shall be equal. Let BCD be the given angle, and ▲ the given point. parallel to CD, make BE=CE, and BE : EC :: BA : AD; but BE-EC; therefore, BA=AD. E |