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(13.) A straight line being drawn parallel to one of the lines containing a given angle and produced to meet the other; through a given point within the angle, to draw a line cutting the other three, so that the part intercepted between the two parallel lines may have a given ratio to the part intercepted between the given point and the other line.

Let ABC be the given angle, DE parallel to AB, and

P the given point.

From P draw PC parallel

B

C

T

A

D

P

to DE or AB, and take BE: CF in the given ratio. Join FP and produce it to A; APF is the line required. For since DE and CP are parallel to AB,

AD: EB :: DF: EF :: PF: CF

..alt. AD: PF:: EB: CF i. e. in the given ratio.

(14.) Two parallel lines being given in position; to draw a third, such that, if from any point in it lines be drawn at given angles to the parallel lines, the intercepted parts may have a given ratio.

Let AB, CD be the given parallel lines; in AB take any point E; and draw EF, EG making angles equal to the given angles; produce EF, and take EH: EG in the given ratio;

L

E

A

K

B

F

G

-D

H

produce FE to I so that FI: IE :: HE : EF; through I draw LI parallel to AB, it is the line required.

Draw IK parallel to EG; then the triangles IEK, EFG are equiangular,

.. IE: IK :: EF: EG

but FI: IE :: HE: EF;

. ex æqu. FI: IK:: HE: EG, i. e. in the given ratio; and IKE=EGF which is one of the given angles, and by construction IFG is equal to the other. Also lines drawn from any point in LI, making with AB and CD angles equal to the given angles will be parallel and equal to FI, IK, and .. in the given ratio.

(15.) If three straight lines drawn from the same point and in the same direction be in continued proportion, and from that point also a line equal to the mean proportional be inclined at any angle; the lines joining the extremity of this line and of the proportionals will contain equal angles.

Let AB AC :: AC: AD, and from A let AE be drawn equal to AC, inclined at any angle to AB; join EB, EC, ED; the angle BEC B is equal to the angle CED.

E

C D

=AC;

For since AB: AC:: AC: AD, and AE =. .. AB : AE :: AE: AD, i. e. the sides about the angle A are proportional, and .. the triangles AEB, AED are similar, and the angle AED is equal to EBA. Also since CA=AE, the angle AEC=ECA; but ECA is equal to the two BEC, EBC, (Eucl. i. 32.) .. also AEC is equal to the two BEC, EBC; of which DEA is equal to EBC; .. the remainder DEC is equal to the remainder BEC.

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GEOMETRICAL PROBLEMS.

(16.) To trisect a right angle.

Let ACB be a right angle. In CA take any point A, and on CA describe an equilateral triangle ACD, and bisect the angle DCA by the straight line CE; the angles BCD, DCE, ECA are equal to one another.

B.

[Sect. 1.

D

E

For the angle DCA being one of the angles of an equilateral triangle is one third of two right angles, and therefore equal to two thirds of a right angle BCA; consequently BCD is one third of BCA; and since the angle DCA is bisected by CE, the angles DCE, ECA are each of them equal to one third of a right angle, and are therefore equal to BCD and to each other.

(17.) To trisect a given finite straight line.

Let AB be the given straight line. On it describe an equilateral triangle ABC; bisect the angles CAB, CBA by the lines AD, BD meeting in D, and draw DE, DF parallel to CA and CB respectively. AB will be trisected in E and F..

E

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Because ED is parallel to AC, the angle EDA = DAC-DAE and therefore AEED. For the same reason DF=FB. But DE being parallel to CA and DF to CB, the angle DEF is equal to the angle CAB, and DFE to CBA, and therefore EDF ACB; and hence the triangle EDF is equiangular, and consequently equilateral; therefore DE EF = FD, and hence AE-EF= FB, and AB is trisected.

=

=

(18.) To divide a given finite straight line into any number of equal parts.

G

L

H

E

I

M

F

Let AB be the given straight line. Let AC be any other indefinite straight line making any angle with AB, and in it take any point D, and take as many lines DE, EF, FC &c. each equal to AD as the number of parts into which AB is to be divided. Join CB, and draw DG, EH, FI &c. parallel to BC; and therefore parallel to each other; and draw DK parallel to AB.

B

C

Then because GD is parallel to HE one of the sides of the triangle AHE, AG GH:: AD: DE; hence AG = GH. For the same reason DL = LM. But DM being parallel to GI, and DG, LH to MI, the figures DH, HM are parallelograms; therefore DL=GH and LM =>HI, consequently GH=HI. In the same Imanner it may i shewn that HI=IB; and so on, there be any other parts; therefore AG, GH, HI, IB, &c. are all equal, and AB is divided as was required.

if

COR. If it be required to divide the line into parts which shall have a given ratio; take AD, DE, EF, &c in the given ratio, and proceed as in the proposition.

(19.) To divide a given finite straight line harmonically.

Let AB be the given straight line. From B draw any straight line BC, and join AC; and from any point. E in AC draw ED parallel to CB, and make FD =

E

D

F G

B

FE, join DC cutting AB in G. AB is harmonically divided in G and F.

Since BC is parallel to FD, the angle BCG is equal to GDF and the vertically opposite angles at G are equal; therefore the triangles DGF, BGC are similar,

and BC: BG :: FD : FG

But FE being parallel to BC,

(Eucl. vi. 2.) AB BC:: AF FE=FD.

:

.. ex æquali, AB:

or AB

:

BG :: AF: FG

AF :: BG: FG.

(20.) If a given finite straight line be harmonically divided, and from its extremities and the points of division lines be drawn to meet in any point, so that those from the extremities of the second proportional may be perpendicular to each other, the line drawn from the extremity of this proportional will bisect the angle formed by the lines drawn from the extremities of the

other two.

Let the straight line AB be divided harmonically in the points G and F, and let the lines AC, BC, GC, FC be drawn to any point C so that GC may be per

E

pendicular to CA, the angle BCF will be bisected by CG. Through G draw EGD parallel to CA meeting CF in D, then EG being parallel to AC, the triangles EGB, ACB are similar; as also the triangles ACF, DFG, hence

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