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

G

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 AGB=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

B

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,

DE, .. IK

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, and AL LB :: IL : LF

and also :: KL : LD,

.. (Eucl. v. 15.) IL: LF :: GL: LH, and (Eucl. v. 19.) IG : HF :: IL : LF. For the same reason, IK: FD :: IL : LF,

.. IG: HF :: IK : 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, AO 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

A

C

INH

G

B

F

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

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

perpendicular to IG; .. FG GI. Also HK = HF, .. IK=GF;

whence FG = GI=IK, and FK is trisected.

(96.) If a straight line be divided into any two parts, and upon the whole line and one of the parts, as diameters, semicircles be described; to determine a point in the less diameter, from which if a perpendicular be drawn cutting the circumferences, and the points of intersection and the extremities of the respective diameters be joined, and these lines produced to meet; the parts of them without the semicircles may have a given ratio.

Let AB be divided into any two parts in the point C, and on AB, AC let semicircles be described. Take AG: AC the duplicate of the given ratio, and make CD: CB :: AG : GB; D will be the point required.

Ꭺ

H

F

K

GL D C

From D draw the perpendicular DFE; join BE, CF, and produce them to H; join AE, CI; and from K draw KL parallel to EA.

Since CD CB :: AG: GB,

comp. and inv. DB: CD: AB AG,

and since CI is parallel to BE, and KL to EA, CD: BE CK :: BA: CL,

BD

:

whence (Eucl. v. 15.) AB : AG :: AB ; CL,
.. AG=CL;

consequently AG: AC: CL: CA,
i. e. in the duplicate ratio of CK: CF,
or (by similar triangles) of HE: HF.

But AG AC is the duplicate of the given ratio,

.. HE HF is in the given ratio.

(97.) If a straight line be divided into any two parts, and from the point of section a perpendicular be erected, which is a mean proportional between one of the parts and the whole line, and a circle described through the extremities of the line and the perpendicular; the whole line, the perpendicular, the aforesaid part, and a perpendicular drawn from its extremity to the circumference will be in continued proportion.

Let AB be divided into any two parts in C, and from C draw the perpendicular CD equal to a mean proportional between AB and AC; and through A, B, D let a circle be de

E

C B

scribed, and draw AE perpendicular to AB; AB, CD, AC, AE are in continued proportion.

=

In AB produced take BF=AC. Join FD, meeting the circumference in G; join AG, AD, GE. Then because BF AC, .. CF AB, and CD is a mean proportional between AC and CF, .. ADG is a right angle, whence (Eucl, iii. 21.) AEG is also a right angle, and equal to EAC; . EG is parallel and equal to AB, i. e. to CF; whence (Eucl. i. 33.) EC and GF are equal and parallel, and the angle ACE CFD= ADC, and the triangles AEC, ADC, CDF are similar,

=

..(CF=) AB : CD :: CD : CA ;; CA : AE.