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(10.) If any number of equal straight lines be placed in a circle; to determine the locus of their points of

bisection.

Let there be any number of lines AB, CD, placed in the circle whose centre is 0, and let them be bisected in E, F; join OE, OF; then (Eucl. iii. 14.) these lines

are equal, and therefore the locus will be a

F

B

circle whose centre is O, and radius equal to the distance of the points of bisection from 0,

(11.) If from a point in the circumference of a circle, any number of chords be drawn; the locus of their points of bisection will be a circle.

From the given point A let any chord AB be drawn in the circle, whose centre is 0; bisect it in D. Join AO, BO and draw DE parallel to BO.

A E

B

Then DE being parallel to BO, the triangles ADE, ABO are similar, and BO is equal AO, .. DE=EA; but AE: AO :: AD : AB (Eucl. vi. 2.), whence AE= AO, .. ED= EA = § AO, and the locus will be a circle described on 40 as a diameter.

(12.) If on the radius of a given semicircle, another semicircle be described, and from the extremity of the diameters any lines be drawn cutting the circumferences, and produced so that the part produced may always have

a given ratio to the part intercepted between the two circumferences; to determine the locus of the extremities of these lines.

On AB the radius of the semicircle AEC let a semicircle ADB be described; and from A draw any line ADE, which produce till EF : ED in the given ratio.

F

E

D

A

B

G

Produce AC to G, making CG: CB, in the given ratio, and join DB, EC, FG;

then since FE: ED ::

:

GC : CB,

.. FE GC :: ED: CB :: DA: AB :: EA: CA, whence (Eucl. vi. 2.) FG is parallel to CE and DB, and the angle AFG is a right angle, and is in a semicircle whose diameter is AG; hence the locus required is a semicircle.

(13.) If from a given point without a given circle, straight lines be drawn, and terminated by the circumference; to determine the locus of the points which divide them in a given ratio.

Let A be the given point and BCD the given circle. Find O its centre and join AO, and divide it in E, so

B

H

that AO : AE in the given ratio, and find a point F, so that EF may be to OD in the given ratio, and with the centre E and radius EF describe a circle; it will be the locus required.

Draw any line AGC; join OC, EG. Since AO : AE in a given ratio, as also OD : EF;

.. OC: EG: AO: AE,

hence OC is parallel to EG,

and AC: AG :: OC : EG, i. e. in the given ratio.

In the same manner it may be shewn that every line drawn from A to BCD will be divided by the circumference of the circle GFH in the same ratio, i. e. GFH will be the locus required.

(14.) Having given the radius of a circle; to determine its centre, when the circle touches two given lines, which are not parallel.

Let BA, AC be the two lines which touch the circle, whose radius is given.

Bisect the angle BAC by the line AE, the centre of the circle will be in this line (Eucl. iv. 4.)

A

B

From A draw AD at right angles to AB and make it equal to the given radius; through D draw DO parallel to AB meeting AE in O; then the centre of the circle being in this line also, must be at the point of intersection O.

(15.) Through three given points which are not in the same straight line, a circle may be described; but no other circle can pass through the same points.

Let A, B, C be the three given points. Join AB, BC, and bisect them in D and E; from which points draw DO, EO at right angles to them; these lines will meet in some point 0; for if not, they are

H

F

D

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B

parallel, and therefore AB, BC must be parallel, which is contrary to the supposition. Join AO, BO, CO.

Since AD=DB, and DO is common, and the angles at D equal, .. AOBO. In the same manner it may be shewn that BO=CO; and the three lines OA, OB, OC being equal, a circle described from the centre O at the distance of any one of them will pass through the extremities of the other two.

And besides this, no other circle can pass through A, B, C: for if it could, its centre would be in DF and EH, and.. in their intersection; but two right lines cut each other only in one point, .. only one circle can be described.

(16.) From two given points on the same side of a line given in position, to draw two straight lines which shall contain a given angle, and be terminated in that line.

Let A and B be the given points

and CD the given line.

Join AB and on it describe a segment

of a circle containing an angle equal to

D

the given angle, and (if the problem be possible) meeting CD in P; P is the point required.

For join PA, PB; the angle APB being in the segment is equal to the given angle.

(17.) If from the extremities of any chord in a circle perpendiculars be drawn, meeting a diameter; the points of intersection are equally distant from the centre.

E

At C and D the extremities of the chord CD, let perpendiculars to it be drawn meeting a diameter AB in E and F; E and F are equally distant from the centre O.

EA

HO

B F

Draw OG perpendicular to CD, and therefore bisecting it; then OG is parallel to DF;

whence GD: OF:: HG: HO:: HC: HE since the triangles HGO, HEC are equiangular; .. (Eucl. v. 18, 15.) DG: OF:: GC: OE but GD GC, .. OF= OE.

=

(18.) If from the extremities of the diameter of a semicircle perpendiculars be let fall on any line cutting the semicircle; the parts intercepted between those perpendiculars and the circumference are equal.

From A and B the extremities of the diameter AB let AC, BD be drawn perpendicular to any line CD cutting the semicircle in E and F; CE is equal to FD.'

From the centre draw OG perpendicular to CD, it will be parallel to AC and BD,

whence CG: GD:: AO: OB, i. e. in a ratio of equality. But (Eucl. iii. 3.) EG=GF, and .. CE=FD.

(19.) In a given circle to place a straight line parallel to a given straight line, and having a given ratio to it.

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