and these are all constant, being subtended by fixed arcs.

.. ARB is constant; and since the points A and B are fixed, the locus is part of a circle. If PA or PB cuts the ce without being produced, the ARB = the supplement of the sum of the Ls P, X, Y. Hence the rest of the circle is obtained.

49. Let

of X.

From the

8

PH and KQ intersect at X.

Required the locus

PXQ it will be seen by 1. 32 that the X = the HPA, AQK; both of which are constant, since they stand on the fixed arcs HA, AK.

diff. of the

And since H, K are fixed points, the locus of X is part of a If P and Q are on the same side of A, the value of the supplementary to that found above, and the rest of the obtained.

.

X is

is

.. the sum of the 4s at the base of AXAB = one-half of the sum of the 4s at the base of ▲ PAB, QAB.

Hence [1. 32] the vertical ▲ AXB = one-half of vertical ▲ 3 APB, AQB, both of which are constant [III. 21].

.. AXB is constant; and A, B are fixed points. .. the locus of X is the arc of a segment of ○ on base AB [III. 21, Cor.].

51. Let C, D be the centres of the two Os, and in the figure considered let X, the middle point of PQ, fall in PA.

Bisect CD at G, and draw CE, GH, DF perp. to PQ.

Then EF = PQ; for EA =

.. EF = XQ: also EH = HF

PA, and AF = 1AQ.

[Ex. 14, p. 98].

Hence it may be shewn that XH = HA.

Then from the ▲ GHX, GHA, we have GX

.. the locus is a circle, with centre G and radius GA or GB. A better proof follows from Book vi., Prop. 6.

Join BP, BX, BQ.

Then for all positions of PQ the angles

of the ▲ BPQ are constant [III. 21 and 1. 32].

.. the ratio BP : PQ is constant [vI. 4]: hence the ratio BP: PX is constant.

But the BPX is constant: hence [vI. 6] the PXB is

constant.

.. the

BXA is constant... the locus of X is the arc of a segment on AB.

MISCELLANEOUS EXAMPLES ON ANGLES IN A CIRCLE.
Page 230.

52. Because the points P, Q, C, B are concyclic;

S

.. the BPQ, BCQ together = two rt. angles [III. 22]. Similarly, the

S

BP'Q', BCQ' together two rt. angles. BP'Q'; .. PQ and P'Q' are par1. [1. 28]. Again, let TAT' be the tangent at A to the circum-.

.. BPQ =

Then

Hence

< TAB = 4 BCA [III. 32].

< BAT'= BPQ [1. 13 and III. 22].

.. TT' is par1. to PQ.

53. [In the fig. taken AB, AC when produced meet the second O at D and E].

Let AT be the tangent at A: then

LTAB=LACB [III. 32]

= BDE [Ex. 5, p. 223];

.. TA is par1. to DE [1. 27].

54. For PTA = 2 TBA [III. 32]; and ATC = 4 CTB;

56. Join AD. Then the points B, F, O, D are concyclic;

57. Let A be the external point, BC the chord of contact, and let the tangent AB be produced to D.

Then BAC=

the diff. of DBC, BCA [1. 32]

= the diff. of ≤ 3 in the alt. segments. [III. 32].

58. Let A be the point of intersection of the two the two diams., and let the line through A meet the C Then in the ▲ AXD, AYE

*, AD, AE

at X and Y.

L DAX = LEAY [Hyp.], and ▲ AXD: = LAYE [III. 31];
.. LADX= LAEY [1. 32].

.. the segments are similar.

59. Let ABX, ABY be the two equal 3, and let the C described from centre A cut the ABY at C and the ABX at D, the points C, D being on the same side of AB.

S

Then the arc AC = the arc AD, for they are cut off from equal ○ by equal chords; and B is a point on the Oce of both of the given ; hence the arcs DA, AC subtend equal angles at B on the same side of AB [III. 27]. That is, BC and BD coincide in direction; or, the points B, C, D are collinear.

60. [In the fig. taken the ▲ ABC is acute angled, and A' is on the minor arc AB].

Because B'A', A'C' are par1. respectively to BA, AC,

.. the A' the A, .. the arc B'C' the arc BC [III. 26].

=

=

61. Join HB, BK, AB.

Then

HBK + LX = ¿ ABK + 2 ABH + LX

= LABK + 2 QPX + X [III. 21]

= LABK+LAQK [1. 32]

= two rt. angles [III. 22];

.. the points H, B, K, X are concyclic [11. 22. Converse].

62. Let AB be the given st. line, P the given point of contact, and X and Y the given points in AB.

[The problem is only possible when P is between X and Y.] At P draw PQ perp. to AB; then the centre of the required lies on PQ.

On XY describe a semicircle, meeting PQ at O.

From centre O, with radius OP, describe a O, and from X and Y draw the tangents XC, YD. These tangents shall be par1. This is proved by shewing by the converse of [Ex. 10, p. 183] that the sum of the CXP, DYP is two rt. angles.

63. Because the

8

..the four points CP [III. 31]. And this a radius of the given O.

8

C,

CXP, CYP are rt. angles,

X, P, Y lie on a whose diameter is is of constant magnitude, since CP is

Now the YCX is also constant. .. the chord XY is constant.

for the points A, N, P, M are obviously concyclic.

Hence MN and PB are par1. [1. 28].

65. Join XN, YN.

8

Then each of the AXN, APB, NYB is a rt. angle [11. 31]. .. the fig. XNPY is a rectangle.

.. 4 NXY = 4 NPY

= NAX, from the rt. angled ▲ PAB, NPB [1. 32]. .. XY touches AXN [Converse of III. 32].

Or, otherwise. Join X to C, the centre of the AXN.
Then CXA = L CAX = L NPB = NXY.

Hence AXN = L CXY.

..CXY is a rt. angle; .. XY is a tangent.

Similarly XY may be proved a tangent to the other circle,

66. Let AB be the common chord, through A draw APXQ to cut the arcs. Then shall PX QX.

For since APB is the suppt. of AQB,

S

..L BPQ = L AQB,

and ▲ BXP, BXQ are rt. angles [III. 31].

Hence PX = QX [1. 26].

67. Let AD, AE be the given lines touching the given at B and C. Let the chord PQ be bisected by BC at Z, and produced to meet AD and AE at X and Y.

Then shall PX = QY.

Take centre O. Join OZ, OB, OC, OX, OY.

Then the OZX, OBX are rt.

s [III. 3, III. 18]; .. the four points O, Z, B, X are concyclic [111. 22]. .. the ZXO = the ZBO, in the same segment.

Similarly, the ZYO = the

But since OB = OC, .. the

ZCO.
ZBO = the

ZCO.

.'. ¿ ZXO` = ▲ ZYO, .'. ZX = ZY [1. 6].

And by hyp. ZP = ZY, .. PX = QY.

68. Let C, D be the centres of the given O3 which intersect at A, and X the given line.

On CD describe a semicircle; and from centre D with radius half of X cut this semicircle at E. Join ED.

Through A draw PAQ par. to ED. PQ shall be the line required. Join CE and produce it to meet PQ at G, and draw DH par1. to CG, meeting PQ at H.

Then since CED is a right [III. 31],

.. CG, DH are perp. to PQ [1. 29], and GH ED.
Also GH is half of PQ [III. 3];

.. PQ = X, and is drawn through A.

69. Let ABC be the given ▲, on the sides of which equilat. ▲ are described externally, and let the ○ about the equilat. As on BC, CA meet at O. Join AO, BO, CÓ.