together be equal to a given line, not greater than the diameter of the circle.

PD at right angles to PB and equal to half the given line; through D draw DE parallel to PB meeting the semicircle in E; join PE; and produce it to C; PC is the line required.

For, draw FG, EH, CI perpendiculars to AB. Join OE; then the angle PEO is a right angle, and .. (Eucl. iii. 3.) EF EC; whence FG and CI together are equal to 2 EH=2 PD= the given line.

=

(48.) If from each extremity of any number of equal adjacent arcs in the circumference of a circle, lines be drawn through two given points in the opposite circumference, and produced till they meet; the angles formed by these lines will be equal.

B

L

K

C

A

H

E

Let AB, BC, be equal arcs, and F, E two points in the opposite circumference, through which let the lines AFI, BEI; BFH, CEH be drawn, so as to meet; the angles at I and H, will be equal.

From E draw EK, EL, respectively parallel to FA,

From A and B the extremities of the diameter AB let tangents AD, BE be drawn, meeting a tangent to any other point C of the circumference, in D and E; and let O be the centre; join DO, EO; the angle DOE is a right angle.

=

Join CO. Then since CE EB, CO=OB, and the angles at C and B, being right angles, are equal, .. the angle CEO=OEB and CEB is bisected by EO. In the same manner it may be shewn that the angle ADC is bisected by DO. And since the angles CEB, CDA are equal to two right angles, .. CDO and CEO are equal to one right angle, and .. (Eucl. i. 32.) DOE is a right angle.

(53.) If from the extremities of the diameter of a circle tangents be drawn; any other tangent to the circle, terminated by them, is so divided at the point of contact, that the radius of the circle is a mean proportional between its segments.

Let AD, BE be two lines touching the circle ABC, (see the last Fig.) at A and B the extremities of its diameter, and meeting DCE any other tangent to the circle; take the centre, and join CO; then will DC : CO: CO CE.

Join DO, EO; then as in the last proposition, it may be shewn that DOE is a right angle; and since from the right angle OC is drawn perpendicular to the base, .. (Eucl. vi. 8.) it is a mean proportional between the segments of the base, or

DC: CO: CO: CE.

(54.) Two circles being given in magnitude and poșition; 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

B

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.

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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

E

B

H

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

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 from 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.