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Through D draw DF parallel to BA. Find O the centre of the circle, join CO, FO. Then AB being parallel to FD, (ii. 1.) AF is equal to BD; and the angle CEA is equal to CDF, i. e. to half the angle COF, which stands on the arc CF equal to CA and BD together.

2. Next, let AB, CD intersect in E, without the circle.

The same construction being made, the angle CEA is equal to the angle CDF, i. e. to half COF, i. e. to half the angle standing on

CF which is the difference between CA and AF or CA and BD.

(25.) If from a point without two circles which do not meet each other, two lines be drawn to their centres, which have the same ratio that their radii have; the angle contained by tangents drawn from that point towards the same parts will be equal to the angle contained by lines drawn to the centres.

From the point A let the

lines AB, AE be drawn to the centres of two circles, and let them have the same ratio that the radii BC, DE, have; from A draw the tangents AC, AD; as also AF, AG; each

of the angles CAD, FAG will be equal to BAE.

E

Since AB BC:: AE: ED and the angles at C and D are right angles, .. the triangles ABC, ADE are

equiangular (Eucl. vi. 7.), and the angle CAB=DAE; to each of these add the angle BAD; and CAD=BAE.

In the same manner FAG may be shewn to be equal to BAD.

(26.) To determine the Arithmetic, Geometric and Harmonic means between two given straight lines.

D

B C

Let AB, BC be the two given lines. Let them be placed in the same straight line, and on AC describe a semicircle ADC. Through B draw BD at right angles to AC, join OD, and upon it let fall the perpendicular BE. Then 40 being half of the sum of AB, BC is the arithmetic mean; and since (Eucl. vi. 8.) AB : BD :: BD : BC, .. BD is the geometric mean. And DE is the harmonic mean, for (Eucl. vi. 8.) (DO=) AO: DB :: DB: DE, i. e. it is a third proportional to the arithmetic and geometric means, and ... is the harmonic mean.

(27.) If on each side of any point in a circle any number of equal arcs be taken, and the extremities of each pair joined; the sum of the chords so drawn, will be equal to the last chord produced to meet a line drawn from the given point through the extremity of the first

arc.

Let AB, BC, CD, &c., AE, EF, FG, &c. be equal arcs and let their extremities BE, CF, DG be joined;

GEOMETRICAL PROBLEMS.

[Sect. 2. from A through B draw ABH meeting the last chord GD in H; GH is equal to EB, CF, DG, &c. together.

B

F

H

Join DF, EC, and produce EC to I. Since AE = BC, ABH is parallel to ECI. And since the arcs are equal, the lines BE, CF, DG are parallel, whence BI is a parallelogram and BE=HI. In the same manner it may be shewn, that CF= ID, and so on, whatever be the number of equal arcs; hence GH is equal to the sum of BE, CF, DG.

(28.) If the circumference of a semicircle be divided into an odd number of equal parts, and through the points which are equally distant from the diameter, lines be drawn; the segments of these lines intercepted be tween radii drawn to the extremities of the most remote, will together be equal to a radius of the circle.

Let the circumference of the semi

circle ADB be divided into
any odd
number of equal parts, e. g. five, (the
demonstration being the same for any
odd number) in the points C, D, E, F.
Join DE, CF, which are parallel, since

D

E

C

L

F

M

A

B

K

Ο

H

they intercept equal arcs. Join OD, OE; DE and LM together are equal to the radius of the circle.

For complete the circle, and divide the opposite semicircle in the same manner; join AC, DG, EH which will be parallel to one another; CH will also be parallel to DI. Hence DE OK, and OK is also equal to each of the two PM, CL, .. PM= CL, whence LM-CP which is equal to AK, since CF is parallel to AB,

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and DE+ LM = KO+ AK = AO the radius of the circle.

(29.) If from the extremities and the point of bisection of any arc of a circle, lines be drawn to any point in the opposite circumference; the sum of those drawn from the extremities will have to that from the point of bisection, the same ratio that the line joining the extremities, has to that joining one of them and the point of bisection.

Let the arc AB be bisected in C, and AB, AC joined; to any point D in the circumference draw AD, BD, CD; then AD +DB : DC :: AB : AC.

Draw AE parallel to CD, and let it meet BD produced in E. The angle EAD ADC = CAB (Eucl. iii. 27.); =

to each of these add DAB, and EAB =

B

CAD; also ABD = ACD, .. the triangles EAB, ACD are equiangular, whence BE CD BA: AC. But since CDB is equal to each of the angles AED, DAE, they are equal to one another, .. DA = DE,

and AD + DB: DC :: AB : AC.

F

(30.) If two equal circles cut each other, and from either point of intersection a circle be described cutting them; the points where this circle cuts them and the other point of intersection of the equal circles are in the same straight line.

Let the two equal circles cut each other in A and B, and with the centre A and any distance AC, describe a circle FCD cutting their circumferences in C and D; C, D, B will be in a straight line.

Join CB, and let it meet the circumference ADB in E. Join AE, AC. Since the angle ABC is an angle

B

F

in each of the two equal circles, the circumference AC is equal to the circumference AE (Eucl. iii. 26.), .. the line AC is equal to the line AE; and .. E is a point in the circle FDC, and being by construction in the circumference ADB, it must coincide with D; .. CB passes through D, or C, D, B are in a straight line.

(31.) If two equal circles cut each other, and from either point of intersection a line be drawn meeting the circumferences; the part of it intercepted between the circumferences will be bisected by the circle whose diameter is the common chord of the equal circles.

Let the two equal circles ADB,

ACB cut each other in A and B ; join AB, and on it as a diameter let

circle AEB be described, and

from A draw any line ADC meet

E

B

ing the circumferences in D and C; DC' is bisected in E.

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