in B and C: shew that the straight lines bisecting the angles ABC, ACB, will always intersect in a third fixed straight line.

35. If the sides AB, AC of a given triangle be cut proportionally in any two points D and F, and perpendiculars to the sides from those points intersect in L: prove that all the points L correspónding to different positions of D and F will be in a straight line passing through A.

36. Show that the locus of the vertices of all the triangles constructed upon a given base, and having their sides in a given ratio, is a circle.

37. On the base of a given isosceles triangle and on the same side of it, describe a triangle having the angle at the vertex double the angle at the vertex of the original triangle. Prove that the vertices of all such triangles lie in a certain circle.

38. Triangles are constructed on the same base, with equal vertical angles; prove that the locus of the centers of the escribed circles, each of which touches one of the sides externally and the other side and base produced, is an arc of a circle, the center of which is on the circumference of the circle circumscribing the triangles.

39. If through any point in the circumference of a circle, chords be drawn and divided, so that the parts are in a constant ratio, the points of division shall all lie on the circumference of another circle.

40. Determine a series of points within a given circle, such that the rectangle contained by the parts of all chords drawn through them shall be equal to the square whose side is half the radius of the circle.

41. Let O, C, be any two fixed points in a circle, and 04 any chord; then if AC be joined and produced to B so that OB may be equal to OA, the locus of B is an equal circle.

42. BCA, PCP' are two straight lines, BCA is fixed, and PCP' revolves about C; shew that the locus of the intersection of BP, P'A is a circle.

43. Find the locus of a point such that the distance between the feet of the perpendiculars from it upon two straight lines given in position may be constant.

44. A ladder is gradually raised against a wall; find the locus of its middle point.

V.

45. If a circle roll within another of twice its size, any point in its circumference will trace out a diameter of the first.

46.

Two segments of a circle are upon the same base AB. Pis any point in one of the segments. The lines APD, BPC are drawn, meeting the other segment in D and C. AC and BD are drawn, intersecting in Q. Shew that the locus of Q is a circle.

47. P, Q are any points in the circumferences of two segments described on the same line AB; the angles PAQ, PBQ are bisected by the lines AR, BR meeting in R: prove that R will always lie on a third segment described on AB.

48. If a rod move betwixt two fixed straight lines CP, CQ; and the perpendiculars from P, Q upon CP, CQ respectively meet in R, whilst those from P, Q upon CQ, CP respectively meet in S: the loci of R and S will be two circles having their common center in C.

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49. If any number of circles be drawn having a common chord AB, and from a point C in AB produced, lines be drawn intersecting the circles; shew that when the locus of the first points of intersection of the straight lines and circles is a circle passing through B and C, then the locus of the second points of intersection is a straight line.

50. If there be two given circles, shew that the locus of points from which tangents drawn to each include a given angle is another circle. 51. Find the locus of a point such that the tangents drawn from it to two fixed circles are equal.

52. A series of circles is described having a common chord, and from a point in the chord produced, tangents are drawn to the circles, show that the points of contact of the tangents with the circles lie in the circumference of another circle.

53. A series of circles touching each other at one point are cut by a fixed circle; shew that the intersections of the pairs of tangents to the latter, at the points where it is cut by each of the other circles, lie in a straight line.

54. Construct the locus of a point whose distance from one given point is double of its distance from another.

55. If a straight line be divided into any two parts, find the locus of the point at which these parts subtend equal angles.

56: Three given straight lines are in the same straight line; find a point from which lines drawn to their extremities shall contain equal angles.

57. If a circle be given in magnitude and position, and a given point either within or without the circle, and if tangents be drawn at the extremities of all the chords which pass through the given point; to find the locus of the intersections of the tangents.

58. If two chords be drawn, one in each of two circles which intersect one another, through any point in the chord which joins the points of intersection of the circles; the locus of the extremities of the two chords is the circumference of a circle.

VI.

59. Two straight lines which intersect and whose point of intersection is fixed, contain a given angle and are always to each other in the same ratio, and the extremity of one of these lines moves on a given straight line; prove that the extremity of the other moves also on a straight line, and determine its position.

60. Find the locus of a point, such that if straight lines be drawn from it to the four corners of a given square, the sum of the squares shall be invariable.

61. If in two parallel lines two points A and B be taken, and through two other points C, D, in the same lines, any pair of parallels be drawn to meet any pair of parallels through the points A, B, in the points E, F; the line passing through the points E, F, shall pass through a point which shall remain fixed as long as A, B, C, D remain fixed. Shew also that if C and D move in a certain manner along the fixed parallels, the line passing through E and F will still pass through the fixed point.

62. If any rectangle be inscribed in a given triangle, required the locus of the point of intersection of its diagonals.

63.

From any point in the base of a triangle lines are drawn parallel to the sides; shew that the intersection of the diagonals of every parallelogram so formed lies in a certain straight line.

From two given points on the same side of a straight line given in posi tion, draw two straight lines which shall meet in that line, and make equal angles with it; also prove, that the sum of these two lines is less than the sum of any other two lines drawn to any other point in the line.

Analysis. Let A, B be the two given points, and CD the given line. Suppose & the required point in the line, such that AG and BG being joined, the angle AGC is equal to the angle BGD.

Draw AF perpendicular to CD and meeting BG produced in E. Then, because the angle BGD is equal to AGF, (hyp.) and also to the vertical angle FGE, (1. 15.)

therefore the angle 4GF is equal to the angle EGF; also the right angle AFG is equal to the right angle EFG, and the side FG is common to the two triangles AFG, EFG, therefore AG is equal to EG, and AF to FE.

Hence the point E being known, the point G is determined by the intersection of CD and BE.

Synthesis. From A draw AF perpendicular to CD, and produce it to E, making FE equal to AF, and join BE cutting CD in G.

Join also AG.

Then AG and BG make equal angles with CD.

For since AF is equal to FE, and FG is common to the two triangles AGF, EGF, and the included angles AFG, EFG are equal; therefore the base AG is equal to the base EG,

and the angle AGF to the angle EGF,

but the angle EGF is equal to the vertical angle BGD,
therefore the angle AGF is equal to the angle BGD;

that is, the straight lines AG and BG make equal angles with the straight line CD.

Also the sum of the lines AG, GB is a minimum.

For take any other point Hin CD, and join EH, HB, AH. Then since any two sides of a triangle are greater than the third side, therefore EH, HB are greater than EB in the triangle EHB.

But EG is equal to. AG, and EH to AH;

therefore AH, HB are greater than AG, GB; and this will always be so, how near soever the point I may be to the point G. If any other point H' be taken on the other side of G, and AH', BH be drawn, it may be shewn in the same way that AH', BH' are always greater than AG, GB, how near soever the point H' may be to G.

But when three consecutive values of a variable magnitude are found, such that the first and third are always greater than the second, the second magnitude has then its least value. Hence the sum of AG, GB is less than the sum of any other two lines which can be drawn from A and B to any other point in CD.

PROPOSITION II. PROBLEM.

From two given points on different sides of a straight line, draw two straight lines to meet at a point in the line, so that their difference shall be the greatest possible.

Let A, B, be the two given points, and CD the given line.

It is required to find a point G in CD, so that the difference of the lines AG, BG drawn from A and B shall be the greatest possible. From B draw BE perpendicular to CD, and produce BE to F, making EF equal to EB.

Join AF and produce it to meet CD in G; then shall be the point required, such that the difference of AG and BG is greater than the difference of any other two lines drawn from A and B to any other point in CD.

Take any other point H in CD, and join AH, HF, HB. Then GF is equal to GB, (Euc. I. 4.) also HF is equal to HB; and the difference of AH, HB is equal to the difference of AH, HF, but in the triangle AHF, the difference of the two sides AH, HF is less than AF, the third side of the triangle.

And AF is the difference of AG and GB;

therefore the difference of AH and HB is always less than AF, the difference of AG and GB, how near soever the point H may be to the point G.

If any other point I be taken on the other side of G, it may be shown in the same manner, that the difference of AH' and H'B is always less than the difference of AG and GB, how near soever the point H' may be to the point G.

And when three consecutive values of a variable magnitude are found, such that the first and third are always less than the second magnitude, the second magnitude has then its greatest value.

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Hence the difference of AG and GB is greater than the difference any other two lines drawn from A and B to any other point in CD.

PROPOSITION III. THEOREM.

Of all triangles having the same base and equal vertical angles, the isosceles triangle is the greatest.

Let ABC be an isosceles triangle having the sides AB, AC equal. Describe a circle about the triangle ABC, and take any point in the arc AC, and join DB, DC.

Then ABC, DBC are two triangles upon the same base BC, and having equal vertical angles.

Through A draw EAF parallel to BC,

produce BD to meet EA in F, and join FC.
Then because BC is parallel to EF,

the triangle BAC is equal to the triangle BFC; (Euc. I. 37.)

but the triangle BFC is greater than the triangle BDC, and will be always greater, so long as the point D does not coincide with the point A, therefore the triangle BAC is always greater than BDC.

If any other point D' be taken in the arc AB, and D'B, DC be drawn, it may be shewn in the same manner that the triangle ABC is always greater than the triangle DBC.

Hence the triangle ABC is greater than any other triangle which has the same base and an equal vertical angle.

If two circles touch each other internally, and from any two points in the circumference of the interior circle, straight lines be drawn to the point of contact, they will contain a greater angle than any other two lines drawn from the same points to any other point in the exterior circumference.

Let the circle PAB touch the circle PDE internally in the point P, and let A, B be any two points in the circumference of the circle PAB, and let AP, BP be drawn to the point of contact P. Then the angle APB shall be greater than the angle contained by any other two lines drawn from A, B to any other point in the circumference of the exterior circle.

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E

Join AB and produce it to meet the exterior circle în D, E. Take any point in the circumference of the exterior circle; and draw BQ intersecting the interior circumference in C, and join CA, QA. The angles APB, ACB are equal, being in the same segment, (Euc. III. 21.) but the angle ACB is greater than the angle AQB, (Euc. 1. 32.) therefore the angle APB is greater than 4QB;

and APB will be always greater than AQB, how near soever the point Q may be to the point P.

If a point & be taken in the arc PE, and Q'A, QB be joined, it may be shown that the angle APB is always greater than the angle AQ'B. Hence the angle APB is a maximum at P, the point of contact.

COR. If a common tangent be drawn to the circles at the point P, and PD, PE be joined, the angle APD is equal to BPE.