(89.) Having given the radii of two circles which cut each other, and the distance of their centres; to draw a straight line of given length through their point of intersection, so as to terminate in their circumferences.

Let the two circles AFD, BGD cut each other in D; on OC the

line joining their centres O and C,

F

E D

' describe a semicircle CEO; and in it, from C place CE equal to half the given line, and through D draw FDG parallel to it; FG will be the line required.

Through E draw OEH, which (Eucl. iii. 31.) will be perpendicular to FG; and draw CI parallel to OH, and.. perpendicular to DG; then (Eucl. iii. 3.) FD and DG are bisected in H and I, and .. FG is double of HI; but HECI being a parallelogram, HI = EC; .. FG is double of EC, and consequently equal to the given line.

(90.) If two circles cut each other; to draw from one of the points of intersection a straight line meeting the circles, so that the part of it intercepted between the circumferences may be equal to a given line.

Let the two circles ABC, ADB cut each other in A and B. Join AB, and draw BC touching the circle ABD. Join AC; and take AF a fourth proportional to BC, BA and the given line; join BF and produce it to E; BFE will be the line required.

D

B

Since the angle AFB together with the angle in the segment ADB or (Eucl. iii. 32.) its equal ABC, are

equal to two right angles, i. e. to the angles AFB, AFE, :. ABC= AFE and ACB=AEF being in the same segment,... the triangles ACB, AEF are equiangular, BC :: AF: FE,

and AB

but AB

BC: AF: the given line,

whence FE is equal to the given line.

(91.) If two circles cut each other; to draw from the point of intersection two lines, the parts of which intercepted between the circumferences may have a given

ratio.

Let the two circles ABC, ABD cut each other in A and B; in the circle ABD place BE, BF which have to each other the given ratio; join AE, AF, and produce AE to G; EG will have to HF the given ratio.

Draw the diameters AC, AD; join

GB, BH, BC, BD; then ADBE being a quadrilateral figure inscribed in a circle, the angles AEB, ADB are equal to two right angles, and .. equal to BEA, BEG, :. BEG=BDA=BFA. And since AGBH is a quadrilateral figure inscribed in a circle, AHB, AGB are equal to two right angles, i. e. to AHB, BHF, .. AGB =BHF; hence the triangles GBE, FBH are equiangular,

.. GE: HF:: BE: BF, i.e. in the given ratio.

(92.) If a semicircle be described on the common chord of two intersecting circles, and a line be drawn

from one extremity of this chord cutting the two circles; the part intercepted between the two shall be divided by the semicircle into segments proportional to perpendiculars drawn in those circles from the other extremity of the chord.

Let the two circles ACB, ADB cut each other in A and B; and on AB, the line joining the points of intersection, as a diameter describe the semicircle AEB, and draw any line AFEG cutting the circumferences in F, E, G, and from B draw BC, BD perpendiculars to AB; then will EF: EG :: BD: BC.

Draw the diameters AC, AD; and

join FB, EB, GB. Then AFBD being a quadrilateral figure inscribed in a circle, the angles AFB, ADB are equal to two right angles, i. e. to AFB, BFE, :. ADB =BFE, and the angle FEB in a semicircle is equal to ABD, whence the triangles FEB, ABD are equiangular, and .. FE: EB: DB: BA.

Again, because the angle 4GB=ACB, and BEG is a right angle, and.. equal to ABC, the triangles EBG, ABC are equiangular, and

(93.) Two circles being given, the circumference of one of which passes through the centre of the other; to

draw a chord from that centre such, that a perpendicular let fall upon it from a given point, may bisect that part of it which is intercepted between the circumferences.

Let O and C be the centres of the two given circles, the circumference of the former passing through C, and let D be the given point. Join CO and produce it both ways to A and B. Join BD, and produce it to E, making

E

G

DE=DB. Draw EF touching the circle AF in F; join CF and produce it to G; and on it let fall the perpendicular DH; then CG is the chord required, and FG is bisected in H.

Draw EI parallel to CG, meeting BG produced in I; produce DH to K. Then BG being perpendicular to CG (Eucl. iii. 31.), is parallel to DHK,

.. BD DE :: IK : KE,

but BD = DE, .. IK=KE, whence FH-HG, and

.. FG is bisected in H.

COR. If it be required to draw CG such, that the perpendicular DH may divide FG in any given ratio, take DE: DB in that ratio, and proceed as in the proposition.

(94.) If any number of circles cut each other in the same points, and from one of these points any number of lines be drawn; the parts of these which are intercepted between the several circumferences have the same ratio.

Let any number of circles ABC, ABE, ABH cut ́each other in the same points A and B, and from A draw

[AGEC, AHFD, meeting the circumferences; then HF: GE:: FD: EC.

Join BG and produce it to K;

then (Eucl. iii. 35.) AL: LB :: LG: LH,

For the same reason, IK: FD:: IL: LF

.. IG: HF:: IK: FD.

FD

In like manner, GE: GI:: GB: GA:: GC: KG:: EC: IK, .. ex æquo GE: HF:: EC: FD.

(95.) In a given circle to place a straight line cutting two radii which are perpendicular to each other, in such a manner that the line itself may be trisected.

Let ABC be the given circle, 40 and OB being two radii at right angles to each other; bisect the angle AOB by OC; at C draw the tangent CD, and make it equal to 3 CO; produce OB to E; join

K

INH

B

OD, and from F draw FGIK parallel to DC; it will be trisected at the points G and 1.

Since the angle at C is a right angle, and COB is half a right angle, .. also CEO is half a right angle, and equal to COE; whence CO-CE. And since HF is parallel to CD,

CE: ED:: HG: GF,

but ED is double of EC, .. FG is double of HG. But HG=HI, since HO bisects the angle IOG, and is