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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 o,
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) par. 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 [111. 16].

4. The radius from the pt. of contact of the inner O is perp
to the tangent [111. 18], and .. bisects the chord of the outer o
[111. 3].

5. The tangents to the inner o are chords of the outer at
equal distances from the centre of the outer; and therefore are
equal chords (111. 14].

Pages 182, 183.


1. Let o be the centre of a o touching AB and AC in B and

Then OB = OC, OA is common, and < $ ABO, ACO are rt. 48
[111. 18], .. in the rt.-angled As AOB, AOC, L OAB = LOAC (Ex.
12, p. 01].

2. Let AO cut BC in D. Then _ BAD = L CAD [Ex. 1]. :. in AS BAD, CAD, BD = DC and LBDA =

= LCDA (1. 4].

3. The chords of the outer which are tangents to the 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 [111. 3].

5. The required locus is the perp. to the given st. line through the given pt. [111. 19].

6. The required locus is the st. line which is par! 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 par. there is no solution unless the given radius is equal to half the perp. distance between the parls

. If they are not par!. 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 o equal to the given st. line. Describe a circle concentric with given o and with radius equal to the distance of the centre from CD. From A draw a tangent to this O. 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 parl. tangents at the extremities of the diameter CAB : and DFE a tangent at F. Then the AS ABE, AFE are identically equal [111. 17 and 1. 8). :. AE bisects 4 BAF. Similarly AD bisects ¿ CAF. :. DAE is a rt. _ [Ex. 2, p. 29].

11. Let ABCD circumscribe a o whose centre is o, the pts. of contact of AB, BC, ... being E, F, G, H. .. AE = AH. [II. 17. Cor.] Similarly BE= BF, DG = DH and CG =CF. :. AE, BE, DG, CG together = AH, DH, BF, CF; i.e. AB, CD together = AD, BC.

12. The opp. sides of a parm, are equal: and the sum of one pair of opp. sides of a quad!. circumscribing a O is equal to the sum of the other pair [Ex. 11]. Hence double of one side = double of the adjacent side. 1. the circumscribing parm, is equilateral. 13. Take fig. of Ex. 11. Then, by 111. 17. Cor. and 1. 8,


Ź DOG = = L DOH; L COG = L COF. ..2 8 AOE, BOE, DOG, COG together = { $ AOH, DOH, BOF, COF;

i.e. Ľ S AOB, COD together = _ $ AOD, BOC.
But these four _ $ together = 4 rt. _ $.
.: L 8 AOB, COD = 8 AOD, BOC = 2 rt. į s.

14. Let o be centre. Then OB, being perp. to tangent BD, is par. to AD. .. [ DAB = LABO. But OA= OB,

..LABO = LOAB. : . AB bisects L CAD.

Draw DF perp.

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

16. See fig. p. 180. Let BCD be given 0, EBF the given diameter produced. Draw BA perp. to EB and equal to the required length. Join AE cutting the o in D. 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 O BCD in D. to ED meeting EB in F. Then LDFE BAE = complement of BEA = half the given angle. Hence the tangents from F contain the angle required (111. 17. Cor.].

Draw DF perp.

18. Let A be the pt. through which the o is to pass.

B the pt. on the given st. line which the o is to touch. Join AB. Draw BC perp. to the given line, and make LBAC = L ABC. Then CA= CB. :: C is the centre of the required O. [See solution of Ex. 28, p. 220.]

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.

20. Describe a O concentric with the given O, having its radius equal to the sum or difference of the radii of the given O and of the required O. A pt. of intersection of the o so described with a line drawn par. to the given line at a distance equal to the radius of the required 0, is the centre of the required o. [See solution to Ex. 33, p. 221.]


Page 186.

1. The sum of _ 8 PAB, PBA is the supplement of the constant LAPB (1. 32], and is therefore constant.

2. The _ QRS, QPS in segment QRPS are equal: and the _ S RQP, RSP in segment RQSP are equal : and the opp.

vertical LS RXQ, SXP are equal. 3. The PBQ is the supplement of the sum of the < s BPQ,

i.e. of s in the segments BPA, BQA of the two Os, which are constant. 4. The < PBX = L PAX (111. 21]= vert. opp. L YAQ




= · YBQ (111. 211 5. The LAOB is the supplement of the sum of the halves of LS PAB, PBA and is . constant [Ex. 1]. Hence locus of o is the arc of a on chord AB. [Converse of Prop. 21, p. 187.]

Page 188.

1. The opp.

LS 8 of a parm. are equal; and if a circle can be described about the parm., they are together equal to two rt. _ *. .. each _ is a rt. L.

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

3. The exterior L = supplement of adjacent interior L= opposite interior L. [See solution to Ex. 5, p. 223.]

Page 190.

1. Let ABCD be a quad! inscribed in a 0. Let BE bisect the int. į at B, and let de bisect the ext. — at D. Then angle CDE = half the supplement of 2 ADC = half the < ABC = L CBE. :. CBDE is concyclic [Conv. of Prop. 21). .. E is on the O ABCD.

2. Let ABC be the A, and P, Q, R any points in the ext. arcs BC, CA, AB. Then the sum of the < $ BAC, BPC = two rt. is [111. 22]. So that the sum of the 4S BAC, BPC, CBA, CQA, ACB, ARB = 6 rt. _S. And of these the _$ BAC, CBA, ACB = ? rt. ¿ $ (1. 32]. .. the _ S BPC, CQA, ARB = 4 rt. <$.

3. Let A be the centre, and AB any radius of the O. With B as centre and BA as radius describe a O cutting the given o in C and D. CD shall divide the given o as required. For 2 CBD in the one segment = L CAD = twice – in the other segment [111. 20).

Page 191.

1. In fig. III. 23, LACB in smaller segment is greater than LADB 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, LAQB > APB [1. 21]. If P is within the segment, AP produced will cut the segment in some pt. Q, so that the ext. LAPB > the int. opp. AQB.

3. Let P and x be on BC, Q on CA. Then QX = QC (Ex. 2, p. 100). LQXC = L QCX =< PRQ, since PCQR is a parm. (Ex. 2, p. 96] .. _QXP is supplement of < PRQ. .. P, R, Q, X are concyclic.

4. If Y and z be the feet of the perps. from B and 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 [111. 10]. Hence the six pts. P, Q, R, X, Y, Z are concyclic.

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