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. (15. 4.) But AC is the fifth part of AG, hence AI is the fifth part of AB. RH F Second. 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 P, CD=Q, DE=R; join 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. Q G For, by reason of the parallels 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. other. Upon DE take join AC; and through E B X A B C the point B, draw BX parallel to AC; DX will be the fourth proportional required: for since BX is parallel to AC, we have DA DB:: DC: DX; now the three first terms of that 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. G 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. BH A D E F For the perpendicular EG, let fall from a point in the circumference upon the diameter, is a mean proportional between DE, DF, the two segments of the diameter; (23.4;) and these segments are equal to the given lines A and B. To divide the given line AB into two parts, such that the greater part shall be a mean proportional between the whole line and the other part. At B the extremity of the line AB, erect the perpendicular BC equal to the half of AB; from the point C as a centre, with the radius CB describe a semi F 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, 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, (30.4,) AE: AB:: AB: AD; hence, by division, (17. 3,) AE-AB: AB :: AB-AD: 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 inversion, (9. 3,) AB AF AF: FB. Scholium. This sort of division of the line AB is called division in extreme and mean ratio. It may farther 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 A, in the given angle BCD, to draw the line BD, so that the segments AB, AD, comprehended between the point A and the two sides of the angle, shall be equal. Through the point A draw AE parallel to CD, make BE-CE, and through the points B and A draw BAD; this will be the line required. For, AE being parallel to CD, we have BE: EC:: BA : AD; BA but BE-EC; therefore BA=AD. (5. 3. Cor. 1.) To describe a square that shall be equivalent to a given parallelogram, or to a given triangle. square described upon XY be equivalent to the parallelogram ABCD. For by construction, AB : XY:: XY: DE; therefore XY2=AB.DE; but AB.DE is the measure of the parallelogram, and XY2 that of the square; consequently they are equivalent. Second. Let ABC be the given trian gle, BC its base, AD its altitude: find x a mean proportional between BC and the half of AD, and let XY be that mean; the square described upon XY B D will be equivalent 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 the given line AD, to describe a rectangle ADEX that shall be equivalent to a given rectangle ABFC. constructed with the lines AD and AX, will be equivalent 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. PROBLEM VIII. B To find a triangle that shall be equivalent to a given polygon. Let ABCDE be the given polygon. Draw first the diagonal CE, cutting off the triangle CDE; through the point D, draw DF parallel to CE, and meeting AE produduced; join CF: the polygon G ABCDE will be equivalent to the polygon ABCF, which has one side less than the original polygon. For the triangles CDE, CFE have the base CE common; they have also the same altitude, since their vertices D, F, are situated in a line DF parallel to the base; these triangles are therefore equivalent. (2. 4. Cor. 2.) Add to each of them the figure ABCE, and there will result the polygon ABCDE equivalent to the polygon ABCF. The angle B may in like manner be cut off, by substituting for the triangle ABC the equivalent triangle AGC, and thus the pentagon ABDE will be changed into an equivalent triangle GCF. The same process may be applied to every other figure; for, by successively diminishing the number of its sides, one being retrenched at each step of the process, the equivalent triangle will at last be found. Scholium. We have already seen that every triano! be changed into an equivalent square; (Prob. 6. |