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draw a tangent BD meeting the side CA produced in D. BD is a mean proportional between AD and DC.

(Eucl. iii. 36.) the rectangle AD, DC is equal to the square of DB; and ... AD: DB:: DB: DC.

(29.) From the obtuse angle of any triangle, to draw a line within the triangle to the opposite side, which shall be a mean proportional between the segments of the side.

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Let ABC be a triangle having the obtuse angle ABC. Describe a circle about it, and produce BA to D, making AD=AB. From D draw DE parallel to AC, meeting the circle in E; join BE, cutting AC in F; BF will be a mean proportional between AF and FC.

For (Eucl. vi. 2.) BF: FE :: BA: AD,

and since BA=AD, .. BF=FE.

Now the rectangle AF, FC is equal to the rectangle BF, FE, i. e. to the square of BF;

.. AF: FB :: FB : FC.

(30.) From the common extremity of the diameters of two semicircles given in magnitude and position; to draw a line meeting the circumferences, so that the rectangle contained by the two chords may be equal to a given square.

Let AB, AC be the diameters drawn from A, and given in magnitude and position. With the centre A,

and radius equal to a side of the given square, describe a circle, cutting the lesser semicircle in D. Draw DE perpendicular to AC, and meeting the other semicircle in F. Join AF, and produce it to G; AG is the line required.

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For joining GC, the triangles AGC, AFE are similar, .. AC AG AF AE,

and the rectangle FA, AG is equal to the rectangle CA, AE, i. e. to the square of AD, which is equal to the given square.

(31.) To draw a line parallel to a given line, which shall be terminated by two others given in position, so as to form with them a triangle equal to a given rectilineal figure.

Let AB, AC be the lines given in position, AD the line to which it is re- E quired to draw a parallel. Describe a rectangular parallelogram AEFG equal to the given figure. Produce EF

GK

to H; and take AK a mean proportional between DH and 2 EF; draw KC parallel to AD; KC is the line required.

For the angles DHA, CAK being equal, as also DAH, ACK, the triangles DAH, AKC are equiangular, and similar; whence

AKC: AHD :: AK2 : DH' :: 2 EF: DH :: 2 EF

Now the rectangle DH, AE is AHD, .. AKC is equal to the to the given rectilineal figure.

[× AE: DH × AE. double of the triangle rectangle EF, AE, i. e.

(32.) To bisect a triangle by a line drawn parallel to one of its sides.

Let ABC be a triangle to be bisected by a line parallel to its side AB. On BC describe a semicircle; bisect BC in O, and draw the perpendicular

E

OD; join CD; and with C as centre, and radius CD, describe a circle cutting CB in E; draw EF parallel to AB; EF bisects the triangle.

(Eucl. vi. 8.) BC : CD :: CD : CO,

.. BC: (CD) CE :: BC: CO :: 2 : 1; but the triangles ABC, FEC are in the duplicate ratio of BC: CE, and.. in the ratio of 2: 1, i. e. EFC is half of ABC, and EF bisects the triangle.

(33.) To divide a given triangle into any number of parts having a given ratio to each other, by lines drawn parallel to one of the sides of the triangle.

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B

M

C

Let ABC be the given triangle; divide AC into parts AE, EF, FC having the same ratio to one another that the parts of the triangle are to have. On AC describe a semicircle, and draw the perpendiculars EG, FH; and with the centre A, and radii AG, AH, describe circles meeting AC in I and K, from which points draw IL, KM parallel to BC; these will divide the triangle in the ratio required.

For the triangles ALI, AKM, ABC are to one another in the duplicate ratio of the sides AI, AK, AC, i. e. in the ratio of the rectangles AC, AE; AC, AF; and the

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.

(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 triangle; from the angle C draw CD parallel to the

given line; and let it be re

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quired to divide the triangle

into five equal parts. On

AD, BD describe semi

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circles AID, BMD; divide AB into five equal parts in the points E, F, G, H; draw El, 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: AN2 :: AD : AE. .. ex æquo, ABC: ANR :: AB : AE :: 5 : 1, i. e. ANR 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.

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

A
P

K

L

F

N M

B

H

I

EG

Q

R

D

C

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 H and I; 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.

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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 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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