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(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.

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 I), and draw DE, DF parallel to CA and CB respectively. AB will be trisected in E and F.

Because ED is parallel to AC, the angle EDA = DAC=DAE and therefore AE=ED. 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.



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

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.

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.

GH= HI. In the same manner it may be shewn that HI=IB; and so on, if there be any other parts; therefore AG, GH, HI, IB, &c. are all equal, and AB is divided as was required.

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 =




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 GDFand 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 : BG :: AF: FG

or AB : 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

perpendicular 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

AF : AC :: FG : DG

but AB, AF :: GB : GF, .. ex æquo AB : AC :: BG : GD.


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But. AB : AC :: GB: GE :: (Eucl. v. 15.) BG : GD :: BG : GE, and therefore GD=GE, and GC is common, and the angles at G are right angles, therefore the angle DCG

GCE, and FCB is bisected by CG.



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(21.) If a straight line be drawn through any point in the line bisecting a given angle, and produced to cut the sides containing that angle, as also a line drawn from the angle perpendicular to the bisecting line; it will be harmonically divided.

Let the angle ABC be bisected by the line BD, and through any point D in this line draw GDFE meeting the sides in G and F, and BE a perpendicular to BD in E; then will EG : EF :: GD : FD.

For through D draw AC parallel to BE and therefore perpendicular to BD, then the angles ADB, CDB being right angles are equal, and ABD = CBD and BD is common to the triangles ADB, CDB, :. AD= DC. But DC being parallel to EB,

EG: GD :: EB: DC :: EB : AD :: EF: FD, since the triangles EFB and AFD are similar,

.. EG : EF:: GD : FD.

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(22.) If from a given point there be drawn three straight lines forming angles less than right angles, and from another given point without them a line be drawn intersecting the others so as to be harmonically divided; then will all lines drawn from that point meeting the three lines be harmonically divided.




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From A let AB, AC, AD be drawn making each of the angles BAC, CAD less than a right angle, and from a given point E let EBD be drawn so as to be harmonically divided in C and B; then will any other line EF be harmonically divided in G and H. Through G draw IK parallel to BD,

then DC : CB :: KG : GI,

But DC: CB :: DE: EB
:. (Eucl. v. 15.) DE : EB :: KG: GI

and alt. DE: KG :: EB: GI and since DE is parallel to GK, (Eucl. vi. 2.)

DE: KG :: EF : FG and EB being parallel to GI,

.:. EB: GI :: EH : HG, whence (Eucl. v. 15.)

EF: FG :: EH : HG and alt. EF: EH :: FG : HG.


(23.) If a straight line be divided into two equal, and also into two unequal parts, and be produced, so that the part produced may have to the whole line so produced the same ratio that the unequal segments of the line have to each other, then shall the distances of the point of unequal section from one extremity of the given bine,

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