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square of AC; or in the ratio of the lines AE, AF, AC; whence ALI, LIKM, MKCB are in the ratio of AE, EF, FC, i. e. in the given ratio.

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(34.) To divide a given triangle into any number of equal parts by lines drawn parallel to a given line. Let ABC be the given 1 tsirr: Jant

08 triangle; from the angle

CM 18 draw CD parallel to the given line; and let it be re

EFTGIT quired to divide the triangle into five equal parts. On AD, BD describe' semicircles AID, B.MD; divide AB into five equal parts in the points E, F, G, H; draw EI, FK, GL, HM

perpendicular to AB; and make AN, AO, AP respectively equal to AI, AK, AL, and BQ=BM; and draw NR, OS, PT, QV, parallel to DC; they divide the triangle as required.

(Eucl. vi. 1.) the triangle ABC : ADC :: AB : AD, (Eucl. vi. 19.) ACD: ANR :: AD : AN' :: AD: AE. :: ex æquo, ABC : ANR :: AB : AE :: 5:1,

i. e. AFR is one fifth of ABC.
In the same manner ABC : AOS :: 5 : 2,

whence NRSO is also one fifth of ABC. And by a similar manner, OPTS and BQV, may each be shewn to be one fifth of ABC, ::. TPQV will also be one fifth of ABC.

Cor. In nearly the same manner the triangle may be divided into any number of parts having a given ratio.

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(35.) To divide a trapezium which has two sides parallel into any number of equal parts, by lines drawn parallel to those sides.

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Let ABCD be the given trapezium having the sides AB, DC parallel. On AB the longer side describe a semicircle AFB; from D draw DE parallel to BC; with the centre B, and radius BE, describe the arc EF, and from F let fall the perpendicular FG; and divide AG into the given number of equal parts, e.g. three, in IH and l; and draw HK, IL at right angles to AB. Make BM, BN respectively equal to BL, BK; and draw MO, NP parallel to BC; and PQ, OR parallel to AB; and produce AD, BC to S.

Since DC = BE= BF, and OR = BM=BL, and PQ=BN=BK, the triangle ORS is to DSC in the duplicate ratio of OR to CD, or of BL to BF, i. e. in the ratio of BI : BG;

whence ODCR : DSC :: GI : GB. In the same manner PDCQ : DSC :: GH: GB,

.:. ODCR : PDCQ :: GI : GH,

and ODCR : PORQ :: GI : IH, i. e: in a given ratio of equality.

And in a similar manner APQB may be shewn to be equal to PORQ. And so on, whatever be the number of equal parts.

Cor. In nearly the same manner, the trapezium might be divided into parts having any given ratio.

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(36.) From one of the angular points of a given square to draw a line meeting one of the opposite sides, and the other produced, in such a manner, that the exterior triangle formed thereby may have a given ratio to

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the square.

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Let ABCD be the given square, and M: N the given ratio. From A to DC (produced if necessary) draw a line AO, such that M: MUN :: DC : A0. With the centre

+N O and radius 0 A, describe a semicircle meeting DC produced in E and F. Join AF; which will be the line required.

Join AE. Then M: M+N :: DC: AO :: ABCD: the rectangle AO, AD. Now the triangle ADE is similar to ABG, and equal to it, since AB = AD; .. the trapezium AECG is equal to ABCD; and the rectangle AO, AD is equal to the triangle AEF, whence M : M+N :: AECG : AEF, ..M:N:: AECG : GCF

:: ABCD : GCF.

(37.) From a given point in the side produced, of a given rectangular parallelogram, to draw a line which shall cut the perpendicular sides and the other side produced, so that the trapesium cut off, which stands on the aforesaid side, may be to the triangle which stands upon the produced part of the opposite side, in a given ratio.

Let AKCD be the given rectangle, and E the given point in the side CD produced. On EC describe a semicircle, and in it place

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EF=ED; joiu FC; and divide EC in G, so that EG : GC in the given ratio, and draw GH at right angles to EC. In AK produced take BK a fourth proportional to EG, GH and FC. Join BE; it is the line required.

For (Eucl. vi. 19.) the triangle ECM: EDI :: EC : ED, :. div. CDIM : ECM :: EC - ED : EC,

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but ECM : BMK :: EC : BK
.:. ex æquo CDIM : BMK :: EC-EF2 : BK?

:: FC?: BK
:: EG? : GH, by construction,

:: EG : GC,
i. e. in the given ratio.

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(38.) Through a given point, between two straight lines containing a given angle, to draw a line which shall cut off a triangle equal to a given figure.

Let AB, AC be the lines containing the given angle BAC, and D the given point. Through D draw DE parallel to AC, and describe a parallelogram EG equal to the given figure.

Draw GH perpendicular to AC, and equal to DE; and make HC

DF; join CD, and produce it to meet AB in B; CB is the line required.

For the triangles EBD, DIF, GIC being similar, are to one another in the duplicate ratio of the sides DE, DF, GC; but the square of HC is equal to the squares of HG, GC; and .. the square of DF is equal to the squares of DE, GC'; whence the triangle DIF is

equal to the triangles DBE, GIC; ;. the triangle ABC is equal to AEFG, i.e. to the given figure.

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(39.) Between two lines given in position, to draw a line equal to a given line, so that the triangle thus formed may be equal to a given rectilineal figure.

Let AB, AC be the lines given in position, and DE the line whose magnitude is given. Bisect it in F, and on DF describe a rectangular parallelogram equal to the given figure. On DE describe a segment of a circle containing an angle equal to the angle at A, and cutting HG in I. Join DI, IE; and make AK = ID, and AL=IE. Join KL; it is the line required.

Since AK=ID, and AL=IE, and the angle at A= DIE, ... KL =

::: KL = DE, and the triangle AKL = IDE = HGFD=the given figure.

(40.) From two given lines to cut off two others, so that the remainder of one may have to the part cut off from the other a given ratio; and the difference of the squares of the other remainder and part cut off from the first may be equal to a given square.

Perpendicular to AB one of the given lines, draw BC equal to a side of the given square; and take AD to the other given line in the given ratio of the part remaining

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