:

all. AO CO: AC: CB, .. (Eucl. v. 15.) DA: DB:: AC: CB.

(66.) Two circles being given in position and magnitude; to draw a straight line cutting them so that the chords in each circle may be equal to a given line, not greater than the diameter of the smaller circle.

Let ABC, EFG be the given circles whose centres are O and M. In each place a line AB, EF, equal to

M

K

АН

the given line; and from the centres draw the perpendiculars OI, MK; and with these distances and centres O and I describe circles which will touch AB, EF in I and K; draw CDGH (ii. 7.) which shall touch these circles in L and N; each of the chords CD and GH will be equal to the given line.

Join OL, MN; these lines are perpendicular to CD and GH, and being respectively equal to OI and MK, CD=AB (Eucl. iii. 14.) and GH-EF; but AB and EF are each equal to the given line, .. CD and GH are also each equal to the given line.

COR. If the intercepted parts are required to have

a given ratio, take AB and EF in that ratio, and make the same construction as in the proposition.

(67.) To determine a point in the arc of a quadrant, through which if a tangent be drawn meeting the sides of the quadrant produced, the intercepted parts may have a given ratio.

Let OA, OB be the sides of a quad- A rant produced; and take M and N two right lines which are in the given ratio, and let OC be a mean proportional between the radius of the quadrant and M,

E

D B

and OD a mean proportional between the radius and N. Join CD, and draw the radius OE cutting it at right angles; E is the point required.

Through E draw the tangent AEB, which being perpendicular to OE (Eucl. iii. 18.), will be parallel to CD, .. AO: OB :: CO: OD,

and since OC and OD are mean proportionals between M and the radius, and N and the radius respectively, M: N in the duplicate ratio of OC : OD,

i. e. in the duplicate ratio of 40 : OB.

But (Eucl. vi. 8. Cor.)

AE: EB in the duplicate ratio of AO : OB, .. AE : EB :: M: N, i. e. in the given ratio.

(68.) If a tangent be drawn to a circle at the extremity of a chord which cuts the diameter at right angles, and from any point in it a perpendicular be let fall; the

segment of the diameter intercepted between that perpendicular and chord is to the intercepted part of the tangent, as the chord is to the diameter.

Let the chord CD be perpendicular to the diameter AB, and let CE touch the circle at C; from any point E in which let EF be drawn perpendicular to AB;

FG: CE: CD: AB.

B

Draw the diameter CH; join HD and draw CI perpendicular to EF. Since EC touches the circle, the angle ECH (Eucl. iii. 18.) is a right angle and .. equal to ICD; whence, taking away from each ICH, the angle ECI=HCD, and EIC, HDC are right angles, .. the triangles ECI, HDC are equiangular,

whence IC CE :: DC: CH,

or GF: CE :: CD: AB.

(69.) If a straight line be placed in a circle, and from its extremities perpendiculars be let fall upon any. diameter; these perpendiculars together will have to the part of the diameter intercepted between them, the same ratio that a line placed in the circle perpendicular to the former line, has to the former line itself.

Let the line CD be placed in the circle ABC, and from its extremities let CE, DF be drawn perpendicular to a diameter AB. From D let DG be drawn perpendicular to DC; then will

CE+DF: EF:: GD: DC.

Join CG, which is therefore a diameter of the circle;

and produce CE to I: join DI, and draw DH perpendicular to CE. Since CI is perpendicular to AB, CE= EI, but HE = DF, .. HI = CE+DF. Now (Eucl. iii. 21.) the angle at G is equal to the angle at I, and CDG, DHI are right angles, .. the triangles CGD, HID are equiangular,

and HI: HD :: DG: DC,

or CE+DF: EF:: DG: DC.

(70.) In a circle to place a straight line of given length, so that perpendiculars drawn to it from two given points in the circumference may have a given ratio.

Let A and B be the given points in the circumference of the circle whose centre is 0. Join BA and produce it, and take AC : CB in the given ratio. In the

H

G

circle place a straight line equal to the given straight line, and from the centre O let fall a perpendicular upon it. With O as centre, and distance equal to this perpendicular describe a circle DG, and from C draw CEDF a tangent to it; then HF is the line required.

For (Eucl. iii. 14.) it is equal to the given straight line. And if from A and B, AE, BI be drawn perpendicular to CF, they are parallel to each other, and the triangles CAE, CBI are similar,

:. AE : BI :: CA : CB, i. e. in the given ratio.

(71.) If from any point in the arc of a segment of a circle a line be drawn perpendicular to the base; and

from the greater segment of the base, and arc, parts be cut off respectively equal to the less; the remaining part of the base shall be equal to the chord of the remaining arc.

From any point B in the arc ABC, let BD be drawn perpendicular to AC; make BF= BC, and DE=DC; join AF; AF will be equal to AE.

EDC

Join FE, EB, FB, BC. Since the arc BC= the arc BF, the straight line BC= BF; and DE being equal to DC, and DB common, and at right angles to EC, .. BE=BC= BF, and the angle BFE is equal to the angle BEF. Now since AFBC is a quadrilateral figure inscribed in a circle, the angles AFB, ACB are equal to two right angles, and .. equal to AEB, CEB, of which ACB = CEB, :. AFB = AEB; but BFE = BEF, consequently AFE=AEF; whence AF-AE.

(72.) If from the point of bisection of any arc of a circle a perpendicular be drawn to the diameter, which passes through one extremity; it will bisect the segment of the chord cut off by the line joining the point of bisection of the arc and the other extremity of the diameter.

Let AC be the arc bisected in D. Join AC, and from D draw DE perpendicular to the diameter AB and meeting AC in G; join BD; AG=GF.

F

D

Because AC is bisected in D, the angle CAD is equal to the angle DBA, i. e. to the angle EDA (Eucl.