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EF. The triangles ACE, DFC, having each a right angle, and the angles at C equal, are equiangular, whence

CE : CF :: AC: CD, i. e. in the giving ratio.

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(9.) From a given point between two indefinite right lines given in position, to draw a line which shall be terminated by the given lines, and bisected in the given point.

Let AB, AC be the given lines, meeting in A. From P the given point draw PD parallel to AC one of the lines, and make DE=DA. Join EP, and produce it to F; then will EF be bisected in P.

For since DP is parallel to AF, (Eucl. vi. 2.) EP : PF :: ED : DA, i. e. in a ratio of equality.

Cor. If it be required to draw a line through P which shall be terminated by the given lines, and divided in any given ratio in P, draw PD parallel to AC, and take AD : DE in the given ratio, and draw EPF, it will be the line required.

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(10.) From a given point without two indefinite right lines given in position ; to draw a line such that the parts intercepted by the point and the lines may

have a given ratio.

Let AB, AC be the given lines, and P the given point. Draw PD parallel to AC, and take AD:DE in the given ratio. Join PE, and produce it to F. Then PF : PE will be in the given ratio.

For the triangles PDE and AEFare similar, having the angles at E equal, as also the angles PDE, EAF, (Eucl. i. 39.)

E

.:. FE : EP :: AE : ED and comp.

PF: PE :: AD : DE, i.e. in the given ratio.

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(11.). From a given point to draw a straight line, which shall cut off from lines containing a given angle, segments that shall have a given ratio.

Let ABC be the given angle, and P the given point, either without or within. In B A take any point A, and take AB: BC in the given ratio. Join AC, and from P draw PDE parallel to AC. PDE is the line required.

For since DE is parallel to AC, (Eucl. vi. 2.) DB : BE :: AB : BC, i.e. in the given ratio.

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(12.) If from a given point any number of straight lines be drawn in a straight line given to position ; to determine the locus of the points of section which divide them in a given ratio.

Let A be the given point, and BC the line given in position. From A draw any line AB, and divide it at E in the given ratio; through E draw EF parallel to BD; it is the locus required.

From A draw any other line AD meeting EF in F; then (Eucl. vi. 2.) AF : FD :: AE : EB, i. e. in the given ratio. In the same manner any other line drawn from A to BD will be divided in the given ratio by EF, which therefore is the locus required.

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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.

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P

Let ABC be the given angle, DE parallel to AB, and P the given point. From P draw PC parallel

A

T 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.

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(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.

PO Let AB, CD be the given parallel I lines; in AB take any point E; and draw EF, EG making angles equal to the given angles; produce EF, an take EH : EG in the given rat produce FE to I so I draw LI parall

Draw IK EFG are

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:. IE: K: I .

but FI : E: I .
.. ex equ. FI : IK :: EE: 15.: 122
and IKE=EGF whici E OIE 1
by construction IFG i ecz: C
drawn from any point in the
angles equal to tbe giver Zero
equal to FI, IK, add :. **

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(15.) If thre: BET 1920
point and in the same üzer:
tion, and from that poina. ?**
proportional be intima

i c CT. IETE
the extremity of the ime :
contain equal angies.

Let AB : ÀC = 40
from A let AE be dam
AC, inclined at any angle
join EB, EC, ED; due angle BBC
is equal to the angle CED
For since AB: ACAC

...ABABE
the angle A are proportiora, za
AED are similar, and the area
Also since CA= AE, the ang
is equal to the

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AEC is en is

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(16.) To trisect a right angle.

Let ACB be a right angle. In CA B; 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.

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(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.

Because ED is parallel to AC, the angle EDA= DACEDAE 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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