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Take the centre of the circle, join DO; then (Eucl. iii. 18.) the angle CDO is a right angle, and .. (Eucl. vi. 8.)

CO: OD :: OD: OE,

or CO OA:: OA : OE,

.. div. and comp. AC: CB :: AE : EB. COR. The converse may easily be proved.

(43.) If from the extremity of the diameter of a given semicircle a straight line be drawn in it, equal to the radius, and from the centre a perpendicular let fall upon it and produced to the circumference; it will be a mean proportional between the lines drawn from the point of intersection with the circumference to the extremities of the diameter.

From B the extremity of the diameter AB let a line BC be drawn, equal to the radius BO; and on it let fall a perpendicular OD meeting the circumference

B

in D; join DB, DA; DO is a mean proportional between DA and DB.

Join DC. Then the angles BAD, BCD on the same base are equal. Also since OD bisects BC, it bisects the arc BDC, .. also the straight line BD=DC and the angle DBC=DCB, but ODA=OAD, .. the triangles ODA, DBC are similar, .. AD : DO :: (BC=) DO : DB.

(44.) If from the extremity of the diameter of a circle, two lines be drawn, one of which cuts a perpen

dicular to the diameter, and the other is drawn to the point where the perpendicular meets the circumference; the latter of these lines is a mean proportional between the cutting line, and that part of it which is intercepted between the perpendicular and the extremity of the diameter.

Let CE be at right angles to the diameter AB of the circle ABC, and from A let AD, AC be drawn, of which AD cuts CE in F, then will

AD AC: AC: AF.

:

D

B

f

E

For since the circumference AE is equal to the circumference AC, (Eucl. iii. 27.) the angle ECA is equal to the angle ADC, and the angle at A is common to the two triangles ADC, ACF, .. the triangles are similar, and

AD: AC:: AC : AF.

But if the point of intersection f be without the circle, draw dH parallel to CG, then, as before, the angle HdA is equal to ACd, and the angle at A common to the triangles AHd, ACd,

:. Ad : AC :: AH : Ad :: AC : Aƒ.

(45.) In the diameter of a circle produced, to determine a point, from which a tangent drawn to the circumference shall be equal to the diameter.

From A the extremity of the diameter AB, draw AD at right angles and equal to AB. Find the centre 0, join OD cutting the circle in C; and through C draw CE at right angles to OD meeting BA produced in E.

B

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.

[blocks in formation]

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 C1 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

A

H

I

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,

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.

=

E

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

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