1

FB. Since EK is parallel to FA, the angle KEB is equal to the angle at I; for the same reason the angle LEC is equal to the angle at H. But since the arcs AB, BC, are equal, and AK, BL being each equal to EF (ii. 1.) are also equal to one another, .. KB, LC, are also equal, and (Eucl. iii. 27.) the angles KEB, LEC, are equal, .. also the angles at I and H are equal. The same may be proved whatever be the number of equal arcs AB, BC.

(49.) To determine a point in the circumference of a circle, from which lines drawn to two other given points, shall have a given ratio.

Let A, B be the two given points; join AB, and divide it in D so that AD: DB may be in the given ratio; bisect the arc ACB in C; join CD, and produce it to E; E is the point required.

Join AE, EB. Since AC=CB, the angle AEC is equal to the angle CEB, .. AB is cut by the line ED bisecting the angle AEB, and consequently (Eucl. vi. 3.) AE : EB :: AD: DB, i. e. in the given ratio.

(50.) If any point be taken in the diameter of a circle, which is not the centre; of all the chords which can be drawn through that point, that is the least which is at right angles to the diameter.

In AB the diameter of the circle ADB, let any point

C be taken which is not the centre, and let DE, FG be any chords drawn through it, of which DE is perpendicular to AB; DE is less than FG.

Take O the centre and draw OH perpendicular to FG. Now in the triangle OCH, the angle at H is a right angle and greater than the angle OCH, .. CO is greater than OH, and consequently (Eucl. iii. 15.) DE is less than FG.

(51.) If from any point without a circle lines be drawn touching it; the angle contained by the tangents, is double the angle contained by the line joining the points of contact and the diameter drawn through one of them. From the point E without the circle ABC let EA, ECD be drawn touching the circle in A and C, and let ED meet the diameter AB, drawn from A, in the point D. Join AC; the angle AEC is double of CAB.

E

B

Through C draw the diameter COF; then the angle FCD is a right angle, and .. equal to EAD, and EDA is common to the triangles EDA, COD, ... the angle COD is equal to AED. But COB is double of CAD, .. AEC is double of CAD.

(52.) If from the extremities of the diameter of a circle tangents be drawn, and produced to intersect atangent to any point of the circumference; the straight lines joining the points of intersection and the centre of the circle form a right angle.

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.

D

B

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 O 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

(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

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. 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= 2 AD ; 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 B and C. Join BC; and from A draw any line AFE; AF : AD :: AD : AE.

B

E

H

From A draw the diameter AG, it will cut BC at right angles in I. 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,

.. AF : AI :: AG : AE. But (Eucl. vi. 8. Cor.) AI: AC :: AC : AG, .. ex æquo, AF: AC: AC: AE,

or AF: AD :: AD : AE.