Then because the angle OAD is equal to OCE, each being a right angle, and the angle at O is common to the two triangles OAD, OCE, and OA=OC, .. AD= CE. But AD was made equal to AB, .. CE=AB, and E is the point required.

(46.) To determine a point in the perpendicular at the extremity of the diameter of a semicircle, from which if a line be drawn to the other extremity of the diameter, the part without the circle may be equal to a given straight line.

From B the extremity of the diameter of the semicircle ADB, let a perpendicular BC be drawn; in which take BE equal to the given line; and

on it as a diameter describe a circle; through the centre of which draw AGF, and with A as centre and radius AF describe a circle cutting BC in C. Join AC; CD is equal to the given line.

Join BD. Then BD being perpendicular to AC,

(Eucl. vi. 8. Cor.) AC: AB :: AB : AD, and (Eucl. iii. 36.) AB : AF :: AG : AB, .. ex æquo, AC: AF :: AG: AD, whence AG AD, and .. DC=GF=BE.

=

(47.) Through a given point without a given circle, to draw a straight line to cut the circle, so that the two perpendiculars drawn from the points of intersection to that diameter which passes through the given point, may

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

Let P be the given point with

out the circle ABC, whose centre D is 0; AB the diameter which

passes through P. scribe a semicircle.

P

AG HOT

On PO de

From P draw

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=2PD=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.

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

B

L

A

K

E

drawn, so as to meet; the angles at I and H, will be equal.

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

FB. Since EK is parallel to FA, the angle KEB is equal to the angle at 1; 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.

B

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 a 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 a tangent to any point of the circumference, the straight lines joining the points of intersection, and the centre of the circle form a right angle.

56

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

A

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

DC: CO:: CO: CE.