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.)

D

A

B

E

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 0.

(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

E

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, .. AO=BO. 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

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 0.

G

E

H O

B

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.

D

G

F

E

B

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.

Let AB be the given line in the circle ABC whose centre is 0. Draw the diameter CD at right angles to AB: and taking a line EF which has to AB the given ratio (Eucl. vi. 12.), place it in the circle ABC; bisect it in G and join OG;

make OH=OG, and through H, draw IK parallel to AB; IK is the line required.

For since OG = OH, .. (Eucl. iii. 14.) IK = EF, and EF: AB in the given ratio; .. IK : AB in the given ratio.

(20.) Through a given point either without or within a given circle, to draw a straight line, the part of which intercepted by the circle, shall be equal to a given line, not greater than the diameter of the circle.

Let P be the given point without the circle ABC, whose centre is O. In the circle place a straight line AB equal to the given straight line; which bisect in E; and join OE. With the

P

F

B

centre O and radius OE describe a circle; this will touch AB in E, since the angles at E are right angles (Eucl. iii. 3.); from P draw PCD touching the circle in F. PCD is the line required.

Join OF. Then OF being equal to OE, CD will be equal to AB (Eucl. iii. 14.), i. e. to the given line.

(21.) From a given point in the diameter of a semicircle produced, to draw a line cutting the semicircle, so

that lines drawn from the points of intersection to the extremities of the diameter, cutting each other, may have a given ratio.

Let C be the given point in the diameter BA produced. Make BC CD in the given ratio; and from the points E and

B

D, in which CD cuts the semicircle, draw EB, AD to the extremities of the diameter. CD is the line required.

Since the angles EDA, EBA in the same segment are equal, and the angle at C common to the two triangles ACD, CEB, the triangles are, equiangular, whence

BE: AD :: BC: CD, i. e. in the given ratio.

(22.) From the circumference of a given circle to draw to a straight line given in position, a line which shall be equal and parallel to a given straight line.

Let AB be the given circle whose centre is O, and DE the line given in position.

D

B

From O draw OF parallel and equal to the given line; and

with the centre F and radius equal to OB, the radius of the given circle describe a circle cutting DE in G: join FG, and draw OA parallel to it; join AG; AG is the line required.

Since FG=OB=OA, and is parallel to it, AG is equal and parallel to OF, and . ́. equal and parallel to the given line.

(23.) The bases of two given circular segments being in the same straight line; to determine a point in it