(54.) Two circles being given in magnitude and position; to find a point in the circumference of one of them, to which if a tangent be drawn cutting the circumference of the other, the part of it intercepted between the two circumferences may be equal to a given line.

Let O and C be the centres of the

two given circles. To any point A in the circumference of one of them let a tangent AB be drawn, and make AB equal to the given line. With the

centre C and distance CB describe a circle DBD cutting the other in the point D, and from D draw DE touching the former given circle; E will be the point required.

=

Join CA, CB, CD, CE. Since CA CE and CB= CD, and the angles at A and E are right angles, .. DE is equal to BA, i. e. to the given line.

If the circle DBD neither cuts nor touches DD, it is evident the problem will be impossible.

(55.) To draw a straight line cutting two concentric circles so that the part of it which is intercepted by the circumference of the greater may be double the part intercepted by the circumference of the less.

Let O be the centre of the two circles. Draw any radius OA of the lesser circle and produce it to B, making AB=AO. On AB describe a semicircle ACB cutting the greater circumference in C; join AC, and produce it to E; CE is the line required.

Join CB; and let fall the perpendicular OD. Then

H

the angle ADO being a right angle is equal to the angle ACB, and the vertically opposite angles at A are equal, and the side OA=AB, .. AC=AD, and DC=2AD; but DC is half of EC and AD half of AF, .. EC is double of AF.

COR. The same construction will apply whatever be the relation required between the two chords. Take OB: OA in the required ratio, and proceed as in the proposition.

(56.) If two circles intersect each other, the centre of the one being in the circumference of the other, and any line be drawn from that centre; the parts of it which are cut off by the common chord, and the two circumferences will be in continued proportion.

From any point A in the circumference of the circle ABG, as a centre and with any radius, let a circle BDC be described, cutting the former in Band C. Join BC, and from A draw any line AFE, AF : AD :: AD : AE.

B

C

From A draw the diameter AG, it will cut BC at right angles in 1. Join GE, AC. The right angle AIF being equal to the right angle AEG, and the angle at A common, the triangles AIF, AEG are similar,

(57.) If a semicircle be described on the side of a quadrant, and from any point in the quadrantal arc u radius be drawn, the part of this radius intercepted between the quadrant and semicircle, is equal to the perpendicular let fall from the same point on their common tangent.

On AB the side of a quadrant let the semicircle AEB be described, and from any point C draw the radius CB, and CD perpendicular to AD a tangent at A; EC=CD.

A

B

E

Join AE, AC; then the angle AEB being in a semicircle, its adjacent angle AEC is a right angle and .. equal to ADC; and BCA=BAC= ACD the alternate angle; .. the two triangles AEC, ACD have two angles in each equal, and one side AC common, .. EC= CD.

COR. Any chord of the semicircle drawn froin the centre of the quadrant, is equal to the perpendicular drawn to the other side from the point in which the chord produced meets the quadrantal arc.

Produce DC to F; then CE being equal to CD, the remainder BE is equal to the remainder CF.

(58.) If a semicircle be described on the side of a quadrant, and a line be drawn from the centre of the quadrant to a common tangent; this line, the parts of it cut off by the circumferences of the quadrant, and of the semicircle, and the segment of the diameter of the semicircle made by a perpendicular from the point where the line meets its circumference, are in continued propor

tion.

On the radius AB of the quadrant AGB let the semicircle AEB be described, and at A draw the tangent AD. From B draw any line BECD meeting the tangent in D, and the circumferences

in C, E; from E let fall the perpendicular EF; then BD, BC, BE, BF are in continued proportion.

Since FE is perpendicular to BA, it is parallel to AD, .. BF: BE :: (BA=) BC : BD,

But (Eucl. vi. 8.) BF: BE :: BE : (BA=) BC, .. (Eucl. v. 15.) also BE: BC :: BC: BD,

and BF: BE :: BE : BC :: BC : BD.

(59.) If the chord of a quadrant be made the diameter of a semicircle, and from its extremities two straight lines be drawn to any point in the circumference of the semicircle; the segment of the greater line intercepted between the two circumferences shall be equal to the less of the two lines.

Let O be the centre of the quadrant ADB, join AB and on it let a semicircle ACB be described; from any point C in which let lines CA, CB be drawn to A and B, of which CB is the greater; CD=CA.

E

Join AD, and complete the circle ABE; take any point E and join EA, EB. Since ADBE is a quadrilateral figure inscribed in a circle, the angles AEB, ADB are equal to two right angles, and .. equal to ADB, ADC; whence AEB=ADC; but AEB is half of AOB which is a right angle, .. ADC is half a right

angle, and ACD being a right angle (Eucl. ii. 31.), CAD is half a right angle, and .. equal to CDA, consequently CA=CD.

(60.) If two circles cut each other so that the circumference of one passes through the centre of the other, and from either point of intersection a straight line be drawn cutting both circumferences; the part intercepted between the two circumferences will be equal to the chord drawn from the other point of intersection to the point where it meets the inner circumference.

Through the centre of the circle ABC, let the circle AOB be described, cutting ABC in A and B. If any line AED be drawn from A, and BE joined; DE will be equal to EB.

ED

B

Draw the diameter AOC; join BC, BD. Then since the angle AOB is equal to AEB, .. the angle COB is equal to DEB. Also the angles OCB, EDB, being in the same segment, are equal to one another, .. the triangles OCB, EDB are equiangular, and .. since OB=OC, the angle OCB is equal to the angle OBC, whence EDB= EBD, and .. ED=EB.

(61.) If from each extremity of the diameter of a circle lines be drawn to any two points in the circumference; the sums of the lines so drawn to each point will have to one another the same ratio that the lines have, which join those points and the opposite extremity of a diameter perpendicular to the former.