Ex. 435.-AD, ST intersect at a point G, such that DG=} AD, SG=ST, and HG = HS.

Hence the join of the ortho- and circum-centres passes through the centroid and mid-centre.

Ex. 436.-The circum-circles of triangles ABC, BCT, CAT, ABT, are equal.

Ex. 437.-T is the in-centre of triangle PQR.

Ex. 438.-A, B, C are the ex-centres of PQR.

Ex. 439.-LABS = LTBC.

L BCS= LTCA.

L CAS=LTAB.

Note that the lines joining S and T to any vertex of the triangle ABC are equally inclined to the internal bisector of the angle at that vertex. Any two lines like AS, AT equally inclined to the bisector of an angle BAC are called with reference to that angle isogonal lines.

Any two points S and T, such that the lines joining them to each angle of a triangle are isogonal with reference to that angle, are called inverse points with reference to the triangle.

Thus the ortho-centre and circum-centre of a triangle are inverse points.

Ex. 440.-One angle of a triangle exceeds another by a right angle. Show that the tangent at one of its vertices to the circum-circle of the triangle is perpendicular to the opposite side.

Ex. 441.—ACB, ADB are two intersecting circles; the tangents at C and D meet in E. Show that B, C, D, E are concyclic. Join AB and use III. 32.

Ex. 442.—ACB, ADB are two intersecting circles; AC, AD are tangents to ADB, ACB at A. Show that ABC= LABD.

Use III. 32.

L

Ex. 443.-AB, BC, CD are three equal chords of a circle ABCD. Show that AB, CD are tangents to the circle passing through B, C and the centre of circle ABCD.

Ex. 444.-AB, AC are tangents to a circle BCD. AB, AC are produced to E and F, so that BE BC=CF. Show that E, B, C, F lie on

a circle whose centre is on circle BCD.

Ex. 445. Two chords of a given circle intersect at right angles at a given point: show that the sum of the squares on the chords is constant. Ex. 446.—If the diagonals AC, BD of a cyclic quadrilateral intersect at right angles at a fixed point P. The mid-points of AB, BC, CD,

lie on a fixed circle whose centre is half-way between P and that of the given circle.

Ex. 447.-Let PA, PD be two straight lines of given length inclined at any angle; in PA take a point B ; find a point C in PD, or PD produced such that rectangle AP, PB shall be equal to rectangle CP, PD. How could you tell by merely considering the angles PAD and PDB whether C falls in PD, at D, or in PD produced?

Ex. 448.-Let two circles touch internally at A, and let the radius of the one be equal to the diameter of the other; draw AB, the diameter of the larger, through A, and BP to touch the smaller circle in P; join AP; show that the square on BP is three times the square on AP.

Ex. 449.-Use III. 15 and III. 35 to show that of all equal rectangles the square has the smallest perimeter.

ON SIMSON'S LINE.

The projections on the sides of a triangle of any point on its circum-circle are in the same straight line.

This straight line is called the 'Simson's line' or 'Pedal line' of the triangle with respect to the given point.

Let O be any point on the circum-circle of triangle ABC, OD, OE, OF perpendiculars from O to BC, CA, AB; then D, E, F are collinear.

B

E

Join OA, OB.

Then. 4s OFB, ODB are rt. 4 s

OFE

the on OB as diamr. would pass through D and F, Simy. the on OA as diamr. would pass through E and F ; hence LOAE in same segt. of O through O, E, A, F, = int. and opp. 4 OBC of cyclic quadl. OBCA; ... Ls OFE, OFD = Ls OFD, OBD

2 rt. LS.

Conversely :-If the projections on the sides of a triangle of a point be in a straight line, that point is on the circum-circle of the triangle. Let the feet D, E, F of the perpr. OD, OE, OF on the sides BC, CA, AB of a ▲ ABC be collinear.

Then O is on the circum o of A ABC.

With the same construction, the on OB as diamr. passes through

D and F,

and that on OA as diamr. through E and F. Hence OBD = rt. OFE of cyclic quad.

= LOAE in same segt. of

OBDF,

through O, E, A, F;

... O, B, C, A are concyclic.

Note that slight modifications of the diagram and demonstrations may be required when O takes a different position with respect to the triangle.

Ex. 450.-If the projections of any vertex of a quadrilateral on the sides and diagonal of the quadrilateral on which it does not lie be collinear, so will also the projections of any other vertex of the quadrilateral on the sides and diagonal on which it does not lie.

Ex. 451. From a point O on the circum-circle of a triangle ABC, any three straight lines OD, OE, OF are drawn, making equal angles with BC, CA, AB, and in the same sense.1 Then D, E, F shall be collinear.

It can easily be shown O, B, D, F are concyclic, and that O, A, E, F are concyclic.

And the demonstration then proceeds as when OD, OE, OF were perpendicular to BC, CA, AB.

Ex. 452.-Enunciate and prove the converse of Ex. 451.

Ex. 453. Find a point whose projections on four given intersecting straight lines shall be collinear.

Ex. 454.-Prove the theorem demonstrated on p. 248 by means of the two theorems just demonstrated on p. 238.

Ex. 455. Show that the Simson's line of O with respect to ABC, and that of A with respect to OBC, are equally inclined to BC.

Generalise this theorem.

Ex. 456.-If the perpendiculars from O to BC, CA, AB meet the circum-circle of ABC in p, q, r, the Simson's line of O will be parallel to each of the lines Ap, Bq, Cr.

Ex. 457.-The Simson-line of a point with respect to an equilateral triangle bisects the radius of the circum-circle drawn to the point.

Ex. 458.-The Simson-line of O bisects the join of O and the ortho-centre of triangle ABC.

Ex. 459.-Circles are described on any three chords OA, OB, OC of a circle as diameters. Shew that their other three points of intersection are in a straight line.

1 I.e. all to the right or all to the left of the lines from O.

ON THE RADICAL AXIS OF TWO CIRCLES.

It has already been stated that there exists an indefinite number of points from which equal tangents can be drawn to two given circles which cut one another, and that all these points lie on a certain straight line (see Ex. 374). This straight line is called the radical axis of the two circles.

We shall now show that the same theorem holds true in general for any two circles, whether they intersect or not; i.e. we shall show that

In general a certain straight line can be drawn such that equal tangents can be drawn to two given circles from an indefinite number of points on it.

Let A and B be the centres of any two circles FGH, CDE.

At any point C on circle CDE not on the line through A, B draw the tangent CP.

Describe a circle CFK touching CP at C and passing through any point F on circle FGH.

Let L be its centre.

[See III. 33.

Then BL passes through C and is at right angles to CP, and AL is at right angles to the common chord or common tangent through F to circles CFK, FGH.

Let this chord or tangent be produced to meet CP at P. Draw PN perpendicular to line through A, B, and PG to touch circle FGH. Join AG.