24. For, with the fig. of p. 225, consider the ▲OBC.

Here BF is the perp. from B on the opp. side CO produced : and CE is the perp. from C on the opp. side BO produced. Now BF and CE intersect in A, and AO produced is perp. to BC. Hence A is the orthocentre of the A OBC.

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25. Consider the circumscribed about the As ABC, OBC; and let X be any point on the Oce of the O BOC, on the side of BC remote from O.

Then the 4

and the

S

BOC, BXC are supplementary [III. 22],

< BOC, BAC are supplementary [Ex. 23, p. 226];

.. 4 BXCL BAC.

Hence the segments BAC, BXC are equal, for they stand on equal bases, and contain equal angles [III. 24], .the circles of which these segments are parts are equal.

26. Consider the ▲ FAB. BD is perp. to the side AF [III. 31], and AE is perp. to BF for the same reason:

.. G, their point of intersection, is the orthocentre of the

▲ AFB.

.. FG (produced, if necessary) is perp. to AB [Ex.

p. 224].

19,

27. It will be seen that D is the orthocentre of the A EAC. For AD, being par1. to BC, would meet EC at rt. angles [1. 29]. And CD, being par1. to AB, would meet EA at rt. angles. Hence ED, produced if necessary, must meet AC at rt. angles [Ex. 19, p. 224].

28. For BCK = ▲ BAK, in same segment

Similarly

= compt. of AKB [III. 31]

= compt. of ACB [III. 21]

= OBC [p. 225, Ex. 20].

KBC = BCO;

.. BO is par1. to KC, and BK par1. to OC [1. 27].

29. For, with the figure of the last exercise, since BOCK is a parTM., .. the diagonals bisect one another [Ex. 5, p. 64]. That is, KO passes through the middle point of BC. Hence the st. line joining O to the middle point of BC, passes through K.

30. For, from Ex. 29, we see that the st. line joining the orthocentre to the middle point of the base passes through an extremity of the diam. drawn from A.

.. ¿APQ is a rt. angle [III. 31]; and since AP is also perp. to BC, .'. PQ is par1. to BC [1. 28].

31. Let SX be the perp. drawn from S the centre of the circum- on BC. Then by [Ex. 29, p. 227] AS and OX meet the ce at the same point Q. And SX, passing through the middle point of AQ, is par1. to AO; .. SX is half of AO [Ex. 3, p. 97].

32. Let S be the centre of the circumscribed about the AABC, and A', B', C' the centres of the Os about the A3 OBC, OCA, OAB.

Then it follows from [Ex. 25, p. 226] that SA' and BC bisect one another at rt. angles. Also SB' and AC.

Hence by [Ex. 31, p. 227] AOA'S Similarly OB = SB'.
Again SA' and AO are par1., for both are perp. to BC.
Similarly SB' and BO are par1... LAOBLA'SB'.

.. A'B' = AB. [1. 4] Similarly B'C' = BC and C'A' = CA. It may be noticed that in the As ABC, A'B'C' the orthocentre of each is the circumcentre of the other.

33. Let AP meet RQ in X. Consider the APRX.

34. Let A be the vertex, O the orthocentre, and S the centre of the circum-O.

From centre S with radius SA describe a O.

Join AO and produce it to meet the ce at G.

Bisect OG at D, and draw the chord BC perp. to AG. Join AB, AC. Then ABC shall be the required A. Proof follows from [Ex. 21, p. 226].

Loci. Page 229.

38. Let BC be the given base, and BAC any ▲ of the system, having the vertical ▲ BAC constant in magnitude, but not fixed in position. Let the bisectors of the exterior angles at B and c intersect at I1.

Then CBI, is half the supplement of the B.

412+4 IBC + 41,CB = two rt. angles [1. 32].

fixed, the locus of I, is the arc of a

.. since the base BC is

segment of a circle [111. 21, Cor.].

NOTE. The locus of in Ex. 36 and the locus of 1 are conjugate arcs of the same O.

39. Let the bisectors meet at X.

Then ▲ PAB, QBA together two rt. angles [1. 29].

=

=

.. ▲ XAB, XBA together one rt. angle [Hyp.].

.. ▲ AXB is a rt. angle [1. 32].

And since AB is fixed, the locus of X is a circle on AB as diameter [III. 31. Converse].

40. Let A be the fixed point, C the centre of the O, and APQ any chord through A, meeting the Oce at P, Q. Let X be the middle point of PQ. Then CX is perp. to PQ [11. 3].

L

That is, the ▲ AXC is a rt. angle, and since AC is a fixed base, the point X lies on the Oce of a on AC as diam.

(i) If A is external, the locus is that part of the C on AC which is intercepted within the given O.

(ii) If A is on the Oce, the locus is a complete described on the radius AC as diam., and having internal contact with the given.

(iii) If A is internal, the locus is a complete falling within the given O.

41. Let A be the given point, and B the common centre of the concentric ". Let P be the point of contact of a tangent from A on any one of these 3. Then APB is a rt. angle [111. 18]. And since A and B are fixed points, the locus is a circle on AB as diam.

42. Let A, B be the fixed points on the e, PQ the arc of constant length but variable position. Let AP, BQ intersect at X. To find the locus of X. [In the fig. taken AP, BQ intersect without the ]. Join PB.

But these are constant angles, being subtended by the constant arcs AB and PQ [III. 21]; .. the X is constant.

the locus is the arc of a segment described on AB [III. 21, Cor.]. When AP, BQ intersect within the O, the value of the X is supplementary to that found above, and the conjugate segment is obtained.

43. Let PA, QB intersect at X. Join PB. PQ and AB do not intersect within the circle, ternal].

[In the fig. taken

and X is also ex

Then X is the diff. of PBQ, XPB [1. 32].
But PBQ is constant, being a rt. angle [III. 31].

Also XPB is constant, being subtended by the fixed arc AB.

.. the X is constant; and since the points A, B are fixed, the locus of X is the arc of a segment [III. 21]. When X is internal, the X is supplementary to the value found above, and the conjugate segment of the locus is obtained.

Or,

BPC is half of BAC, and is therefore constant.

Then, since BC is fixed, the locus of P is the arc of a segment on BC [III. 21, Cor.].

45. The intersection of the diagonals is X, the middle point of BC [Ex. 5, p. 64]. Join X to D, the middle point of AB. Then XD is par1. to AC [Ex. 2, p. 96].

.. LDXB = LACB [1. 29].

But ACB is constant [III. 21].

.. 4 DXB is constant, and D, B are fixed points.

.. the locus of X is a O, the segment on DB being similar to the segment ACB.

46. Let A be the point of intersection of the rulers.

Then PXQA is a rectangle.

..AX = PQ, which is constant, and the point A is fixed. Hence the locus of X is the quadrant of a circle described from the centre A with radius PQ.

47. Prove as in [Ex. 9, p. 216] that PXQ=▲ CAD, (or is supplementary to it).

But C, A, D are fixed points; and the arms PX, QX pass through two fixed points C, D.

.. the locus is a

And since

through C and D.

CBD = CAD [1. 8],

.. this passes through B.

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48. [Take the figure in which PA and PB must both be produced to meet the second ce.] Let AY, BX intersect at R. Then the locus of R is required.