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2. Let XZY, X'Z'Y' be two chords bisected at Z and Z' in
AB, of which Z is nearer than Z' to C the middle pt. of AB. Let
O be the centre. Then OC, OZ, Oz' are respectively perp. to AB,
XY, X'Y'. .. oz'> oz > oc [Ex. 3, p. 93]. .. X'Y' <XY < AB.
Hence AB is the greatest length of XY, and XY increases as Z
approaches C. When Z coincides with A or B, XY vanishes.

3. Place any chord PQ of required length in the O. [See
solution of Ex. 4, p. 173 or iv. 1.] Let O be centre of given 0,
and AB the given chord upon which the middle pt. of the required
chord is to lie. Draw ON perp. to PQ. With centre O and radius
ON describe a circle cutting AB in Z and Z'. Then the chord XZY
perp. to OZ will be equal to PQ, and be bisected at Z in AB.
There is no solution if PQ > AB [Ex. 2]; one solution if PQ = AB;
and two solutions if PQ < AB.

Page 181.

1. Draw a diameter (i) at right angles to, (ii) par1. to the
given straight line. At either extremity of the diameter draw
a line perp. to the diameter. These will be the tangents, (i) and
(ii), required.

2. The tangents are perp. to the same diameter, and there-
fore parallel [1. 28].

3. The pt. of contact is in the line of centres [111. 11, 12],
.. the st. line drawn from the pt. of contact perp. to the line of
centres is a tangent to both circles [III. 16].

4. The radius from the pt. of contact of the inner is perp.
to the tangent [III. 18], and .. bisects the chord of the outer
[III. 3].

5. The tangents to the inner are chords of the outer at
equal distances from the centre of the outer; and therefore are
equal chords [11. 14].

Pages 182, 183.

1. Let O be the centre of a touching AB and AC in B and
C. Then OB = OC, OA is common, and $ ABO, ACO are rt. 43
[III. 18],.. in the rt.-angled ▲ AOB, AOC, OAB = ▲ OAC [Ex.
12, p. 91].

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which are tangents to the

3. The chords of the outer inner are equal [Ex. 5, p. 181] and are bisected at the pt. of contact [Ex. 4, p. 181]. Hence the tangents, that is the halfchords, are equal.

4. The tangent at an extremity of a diameter is perp. to the diameter. .. the chords par. to it are bisected by the

diameter [III. 3].

5. The required locus is the perp. to the given st. line through the given pt. [II. 19].

6. The required locus is the st. line which is par1. to the two given st. lines and equidistant from them.

7. The required locus is the pair of bisectors of the angles between the two given st. lines [Ex. 1, p. 182].

8. If the lines are par1. there is no solution unless the given radius is equal to half the perp. distance between the parls. If they are not par1. let them be OX, OY; at O draw OA, OB equal to given radius and perp. to OX, OY respectively. Through A and B draw AP, BP parl. to OX, OY respectively; then P is the centre of the required O.

9. Let A be given pt. Place a chord CD in the given equal to the given st. line. Describe a circle concentric with given and with radius equal to the distance of the centre from CD. From A draw a tangent to this . The tangent is the required chord.

If A is without the circle, the given line must be not greater than the diameter. If A is within the circle, the given line must also be not less than the chord through A perp. to the line joining A to the centre.

10. Let CD, BE be the two par1. tangents of the diameter CAB: and DFE a tangent at F. AFE are identically equal [III. 17 and 1. 8]. .. Similarly AD bisects CAF. .. DAE is a rt.

at the extremities Then the ▲ ABE, AE bisects BAF. [Ex. 2, p. 29].

.'. AE = AH. [III. 17.

11. Let ABCD circumscribe a whose centre is O, the pts. of contact of AB, BC, ... being E, F, G, H. Cor.] Similarly BE = BF, DG DG, CG together

BC.

12.

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DH and CG = CF. .. AE, BE,

AH, DH, BF, CF; i.e. AB, CD together

= AD,

The opp. sides of a parm. are equal: and the sum of one pair of opp. sides of a quad'. circumscribing a is equal to the sum of the other pair [Ex. 11]. Hence double of one side double of the adjacent side. . the circumscribing parm. is equilateral.

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13. Take fig. of Ex. 11. Then, by III. 17. Cor. and 1. 8, LAOE = LAOH; L BOE = L BOF;

▲ DOG = DOH; COG = L COF.

.'. ▲a AOE, BOE, DOG, COG together =LS AOH, DOH, BOF, COF; i.e. s AOB, COD together: =LS AOD, BOC.

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.. ▲ AOB, COD = ▲ AOD, BOC = 2 rt. §.

Let o be centre.

14. is par1. to AD.

..

Then OB, being perp. to tangent BD, DAB = LABO. But OA = OB,

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15. Let O be centre, and let AT, BT' be two equal tangents at A and B. Then OTOT' [III. 18 and 1. 4]. .. locus of T is circle with centre O.

16. See fig. p. 180. Let BCD be given O, EBF the given diameter produced. Draw BA perp. to EB and equal to the required length. Join AE cutting the in D. Draw DF perp. to ED meeting EBF in F. Then the tangent DF = AB, the given length. Hence F is the required pt.

17. See fig. p. 180. At E the centre make the angle BEA equal to the complement of half the given angle. Let EA meet the tangent at B in A, and the BCD in D. Draw DF perp.

to ED meeting EB in F. Then DFE = BAE = complement of BEA = half the given angle. Hence the tangents from F contain the angle required [III. 17. Cor.].

B the

Draw

18. Let A be the pt. through which the is to pass. pt. on the given st. line which the O is to touch. Join AB. BC perp. to the given line, and make BAC = ABC. CA = CB. .. C is the centre of the required . [See solution of

Ex. 28, p. 220.]

Then

19. Let the line drawn parallel to the 'tangent' line at a distance from it equal to the given radius cut the 'centre' line in Then O is the required centre. Two solutions.

O.

SO

20. Describe a concentric with the given O, having its radius equal to the sum or difference of the radii of the given and of the required. A pt. of intersection of the described with a line drawn par1. to the given line at a distance equal to the radius of the required O, is the centre of the required O. [See solution to Ex. 33, p. 221.]

Page 186.

1.

8

The sum of PAB, PBA is the supplement of the constant APB [1. 32], and is therefore constant.

2.

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The QRS, QPS in segment QRPS are equal: and the L RQP, RSP in segment RQSP are equal: and the opp. vertical < RXQ, SXP are equal.

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S

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3. The PBQ is the supplement of the sum of the BPQ, BQP, i.e. of in the segments BPA, BQA of the two ", which are constant.

S

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4. The PBX: = PAX [III. 21] vert. opp. 4 YAQ

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<3 PAB, PBA the arc of a

=YBQ [III. 21].

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AOB is the supplement of the sum of the halves of and is .. constant [Ex. 1]. Hence locus of O is on chord AB. [Converse of Prop. 21, p. 187.]

8

Page 188.

1. The opp. 4a of a parm. are equal; and if a circle can be described about the parm., they are together equal to two rt. ≤3. .. each is a rt. .

2. The ABC = LAXY = < AYX = supplement of XYC. .. XBCY is concyclic. [See Converse of Prop. 22, p. 189.]

3. The exterior

posite interior ▲.

=

supplement of adjacent interior = op[See solution to Ex. 5, p. 223.]

Page 190.

Let BE bisect
Then angle

1. Let ABCD be a quad'. inscribed in a ☺. the int. at B, and let DE bisect the ext. at D. CDE = half the supplement of ADC half the ABC = .. CBDE is concyclic [Conv. of Prop. 21]... E is on the

2. Let ABC be the ▲, and P, Q, R arcs BC, CA, AB. Then the sum of the [III. 22]. So that the sum of the ACB, ARB = 6 rt. ▲ 3. rt. 1. 32].

L

3.

.. the

S

▲ CBE. ABCD.

any points in the ext.

" BAC, BPC = two rt. BAC, BPC, CBA, CQA, And of these the BAC, CBA, ACB=2 s BPC, CQA, ARB = 4 rt. S.

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in

Let A be the centre, and AB any radius of the ☺. With B as centre and BA as radius describe a cutting the given C and D. CD shall divide the given as required. For CBD in the one segment ▲ CAD = twice in the other segment [111. 20].

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Page 191.

1. In fig. III. 23, ACB in smaller segment is greater than ▲ ADB in larger segment [1. 16].

2. If P is without the segment, some part of the arc of the segment must lie within the APB. If Q is any pt. on this part of the arc, AQB > APB [1. 21]. If P is within the segment, AP produced will cut the segment in some pt. Q, so that the ext. APB > the int. opp. ▲ AQB.

=

3. Let P and X be on BC, Q on CA. Then QX QC [Ex. 2, .. _ QXC = 4 QCX = ▲ PRQ, since PCQR is a parm. [Ex. 2, ..QXP is supplement of PRQ. .. P, R, Q, X are

p. 100].

p. 96].

concyclic.

4. If Y and Z be the feet of the perps. from B and C on CA and AB, P, Q, R, Y and P, Q, R, Z are concyclic. But only one circle can pass through P, Q, R [III. 10]. Hence the six pts. P, Q, R, X, Y, Z are concyclic.

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