Ex. 45. Two circles A and B are orthogonal if the tangent to A from the centre of B is equal to the radius of B.

Ex. 46. The tangents drawn from a point P to two circles are equal; prove that a circle can be described with P as centre to cut both circles orthogonally.

Ex. 47. Two circles intersect at O and the tangents at O to the circles are at right angles. The tangent to the first circle cuts the second circle at P and the tangent to the second circle cuts the first circle at Q: show that PQ passes through the other point of intersection of the circles.

Ex. 48. Describe a circle to cut a given circle orthogonally at a given point and to pass through another given point.

RADICAL Axes.

Ex. 49. If two circles intersect, prove that tangents to the two circles from any point on their common chord produced are equal.

Ex. 50. Two circles, centres A, B, and radii a, b, do not intersect. P is a point from which the tangents to the two circles are equal and PN is perpendicular to the line AB. Prove that

AN2-BN2 a2 – b2.

Ex. 51. Prove that the locus of a point from which tangents to two fixed circles are equal is a fixed straight line (called the radical axis).

Ex. 52. Two given circles are cut by a third circle in P, Q and R, S and the chords PQ and RS are produced to meet in T. Prove that T is a point on the radical axis of the two given circles.

Hence, give a geometric construction for the radical axis of two circles which do not intersect.

Ex. 53. Prove that the three radical axes of three circles, taken in pairs, are concurrent.

INVERSION.

If O is a fixed point and P any other point, and if a point P' is taken in OP (produced if necessary) such that OP. OP' = k2, P and P' are said to be the inverse points with respect to a circle whose centre is O and radius k. Sometimes P and P' are said to be inverse points with respect to O.

Ex. 54. O is a fixed point and P is any point in a fixed straight line AB. Prove that the locus of a point Q in OP, such that OP. OQ=constant, is a circle. [Draw ON perpendicular to AB; in ON find a point C such that OC. ON=the given constant; join CQ.]

Ex. 55. O is a fixed point on a circle and P a variable point on the circle; in OP produced take a point P' such that OP. OP'a constant. Prove that the locus of P' is a straight line.

SIMILITUDE.

Ex. 56. Prove that if a point O is joined to each vertex of a given polygon and each join is divided in the same ratio, these points of division are the vertices of a similar polygon.

Extend this principle to the case in which a point P moving round any figure is joined to O.

Ex. 57. Prove that if any two similar triangles ABC, A'B'C' are so placed that corresponding sides are parallel, the joins of corresponding vertices are concurrent.

[Let AA', BB' meet at O. What is ratio OA:OA'? Show that CC' passes through this point.]

Ex. 58. Extend Ex. 57 for two similar rectilinear figures.

Figures placed as in Ex. 57 are said to be similar and similarly situated and the point O is a centre of similitude.

Ex. 59. Draw three figures showing two similar and similarly situated triangles ABC, A'B'C' with their centre of similitude (i) inside both triangles, (ii) outside both triangles and not between A and A', (iii) outside both triangles and between A and A'.

Ex. 60. Prove that for two circles which do not intersect there are two centres of similitude, viz. the points which divide the line joining the centres externally and internally in the ratio of the radii.

Ex. 61. If two circles have real common tangents, these meet the line of centres of the circles at the centres of similitude.

Ex. 62. If a line be drawn from S a centre of similitude of two circles to cut them in P, Q and P', Q' respectively, prove that SQ. SP'= SP. SQ'a constant.

Ex. 63. If a figure is rotated about a point O through any angle a, the angle through which any line in the figure has been rotated is a.

Ex. 64. O is a fixed point and A, B two variable points so related that the triangle OAB is always equiangular with a given triangle; show that, if A moves so as to describe a polygon, B will move so as to describe a similar polygon. [A' in OA, such that OA' : OA= OB: OA, will describe a similar polygon. Now use Ex. 63.]

SOME CONSTRUCTIONS EFFECTED WITH THE AID OF A CENTRE

OF SIMILITUDE.

Ex. 65. Construct a polygon similar to a given polygon without using compasses, a graduated ruler or a protractor.

Ex. 66. Repeat Ex. 65, the required polygon being described on a given line parallel to the corresponding side of the given polygon.

Ex. 67. Inscribe a square in an equilateral triangle ABC with two vertices on BC and one on each of AB, AC.

[Draw a square on BC below the ▲; use A as centre of similitude.]

Ex. 68. Repeat Ex. 67 for any triangle ABC.

Ex. 69. Inscribe a square in a sector of a circle with two vertices on the arc and one on each of the radii.

Ex. 70. Inscribe a square in a semicircle with two vertices on the diameter and the other two on the circumference.

[Draw a square on the diameter and use centre of circle as centre of similitude.]

Ex. 71. Find points D, E, F in the sides BC, CA, AB of a given triangle such that the angles of the triangle DEF are given.

Ex. 72. Inscribe a square in a regular pentagon with one side parallel to a side of the pentagon and one of its four vertices on each of the other sides of the pentagon.

Ex. 73. Show how to describe a circle to touch two given straight lines and pass through a given point.

Ex. 74. Show how to inscribe in a given triangle a triangle which has its sides parallel to the sides of another given triangle.

Ex. 75. Show how to inscribe an equilateral triangle in a given triangle. (There are an infinite number of solutions.)

Ex. 76. Draw any triangle ABC. It is required to inscribe in this triangle an equilateral triangle one side of which is parallel to AB, and the opposite vertex lies on AB.

Show how this can be done by employing the properties of similar triangles.

CONSTRUCTION OF REGULAR PENTAGON.

To divide a given straight line into two parts such that the square on the greater part may be equal to the rectangle contained by the whole line and the smaller part.

[Analysis. Let the whole line contain a units of length.

Let the ratio of the greater part to the whole line be x: 1. Then the greater part contains ax units; and the smaller a units.

ax

The square on the greater part contains a2x2 units of area and the rectangle contained by the whole line and the smaller part contains a (a - ax) units of area,

.. a2x2 = a2 — a2x,

For the present* we reject the lower sign, which would give a negative value for x; and we are left with

In order to construct this length with ungraduated ruler and compasses only, we proceed as follows:—

Let AB be the given straight line.

Construction At A erect ACL to AB, and equal

to AB. Join CB. From CB cut off CD = CA.
From BA cut off BE BD. Then AB is divided
as required.

Proof

=

To verify that this length satisfies the given conditions.

Fig. 139.

Extreme and mean ratio. The relation AE. AB = BE2

=

may be written AE: BE BE: AB. Thus the straight line AB has been divided so that the larger part is the mean proportional between the smaller part and the whole line. In other words, the larger part is the mean, while the smaller part and the whole line are the extremes of a proportion. For this reason, a line divided as above is said to be divided in extreme and mean ratio. This method of dividing a line is also known as medial section.

* It will be seen below (p. 144) that a meaning can be found for the negative value of x.

NOTE. The solution x =

√√5_1

2 2

was rejected. Strictly speaking,

however, it is a second solution of the problem. The fact that this value of x is negative indicates that BE must be measured from B in the other direction-away from A rather than towards A-as BE' in fig. 140.

Ex. 77. With ruler and compass, divide a straight line one decimetre long in extreme and mean ratio. Calculate the correct lengths for the two parts, and estimate the percentage error in your drawing.

Ex. 78. Devise a geometrical construction for dividing a line externally as in the above Note (fig. 140).

Ex. 79. Prove that, if E' is constructed as in the Note (fig. 140), then AB.AE' = BE"; and hence that the line AB is divided externally in extreme and mean ratio.

Ex. 80. Prove that if AB is divided externally in extreme and mean ratio at E', then AE' is divided internally in extreme and mean ratio at B.

Ex. 81. Show how to divide a straight line AB at C so that

(i) AB. AC=2CB2, (ii) 2AB. AC=CB2, (iii) AC2=2CB2.

To construct an isosceles triangle such that each of the base angles is twice the vertical angle. Construction Draw a straight line AB of

any length. Divide AB at C so that AB. BC= AC2. With centre A and radius AB describe a circle. In this circle place a chord BD = AC. Join AD. Then ABD is an isosceles ▲ having ▲ B = L D = 2 LA. Proof Join CD.

Since BC. BA = AC2 = BD2,

.. BC: BD = BD : BA.

Thus, in the As BCD, BDA, the ▲ B is common and the sides about the common

angle are proportional.

.. the As are similar.

Th. 56.

Fig. 141.

But ▲ BDA is isosceles (... AB = AD), .. ▲ BCD is isosceles,

... CD

BD = CA. ... L CDA = LA.

Now BCD (ext. 4 of ACAD) = LA+LCDA = 2 LA,

.. LB=2L A.