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. D B I C 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 we shall have AD.BC+AB.CD=AC.BD. G 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. A CAI DK 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 Estraight line GB; then draw CI parallel to F GB: AI will be the fifth part of the line AB; and thus, by applying Al five times upon AB, the line AB will be divided into five equal parts. G For, since CI is parallel to GB, the sides AG, AB, are cut proportionally in Cand I (Prop. XV.). But AC is the fifth part of AG, hence AI is the fifth part of AB, A 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 P, CD=Q, DE=R; join R the extremities E and B; and through the points C, C IF B D ENG 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 paral.els 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. PROBLEM II. To find a fourth proportional to three given lines, A, B, C. nite lines DE, DF, form- B A parallel to AC; DX will be the fourth proportional required; for, since BX is parallel to AC, we have the proportion DA: DB :: DC: 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; upon the whole line DF, as a diameter, describe the semicircle DGF; at the point E, erect upon the diameter the perpendicular EG meeting the circumference 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 DE, EF, the two segments of the diameter (Prop. XXIII. Cor.); and these segments are equal to the given lines A and B. PROBLEM IV. To divide a given line into two parts, such that the greater part shall be a mean proportional between the whole line and the other part. 155356 the line AB will be divided at the point F in the manner required; that is, we shall have AB : AF :: AF : FB. For, AB being perpendicular to the radius at its extremity, is a tangent; and if AC be produced till it again meets the circumference in E, we shall have AE : AB :: AB: AD (Prop. XXX.); hence, by division, AE-AB: AB :: ABAD: AD. 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 exchanging the extremes for the means, AB : AF :: AF : FB. Scholium. This sort of division of the line AB is called division in extreme and mean ratio: the use of it will be perceived in a future part of the work. It may further be observed, that the secant AE is divided in extreme and mean ratio at the point D; for, since AB=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 A the given point. Through the point A, draw AE paral C lel to CD, make BE-CE, and through the points B and A draw BAD; this will be the line required. E For, AE being parallel to CD, we have BE: EC :: BA: AD; but BE=FC; therefore BA=AD. PROBLEM VI. To describe a square that shall be equivalent to a given parallelogram, or to a given triangle. XY be equivalent to the parallelogram ABCD. For, by construction, AB : XY :: XY : DE; therefore, XY=AB.DE; but AB.DE is the measure of the parallelogram, and XY2 that of the square; consequently, they are equiva lent. valent to the triangle ABC. For, since BC: XY :: XY : AD, it follows that XY2= BC.AD; hence the square described upon XY is equivalent to the triangle ABC. PROBLEM VII. Upon a given line, to describe a rectangle that shall be equivalent to a given rectangle. Let AD be the line, and ABFC the given rectangle. valent to the rectangle ABFC. For, since AD : AB :: AC: AX, it follows that AD.AX= AB.AC; hence the rectangle ADEX is equivalent to the rectangle ABFC. |