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ORTHOGONAL CIRCLES.

DEFINITION. Circles which intersect at a point, so that the two tangents at that point are at right angles to one another, are said to be orthogonal, or to cut one another orthogonally.

39. In two intersecting circles the angle between the tangents at one point of intersection is equal to the angle between the tangents at the other.

40. If two circles cut one another orthogonally, the tangent to each circle at a point of intersection will pass through the centre of the other circle.

41. If two circles cut one another orthogonally, the square on the distance Between their centres is equal to the sum of the squares on their radii.

42. Find the locus of the centres of all circles which cut a given circle orthogonally at a given point.

43. Describe a circle to pass through a given point and cut a given circle orthogonally at a given point.

III.

ON ANGLES IN SEGMENTS, AND ANGLES AT THE
CENTRES AND CIRCUMFERENCES OF CIRCLES.

[See Propositions 20, 21, 22; 26, 27, 28, 29, 31, 32, 33, 34.]

1. If two chords intersect within a circle, they form an angle equal to that at the centre, subtended by half the sum of the arcs they cut off

Let AB and CD be two chords, intersecting at E within the given O ADBC. Then shall the L AEC be equal to the angle at the centre, subtended by half the sum of the

B arcs AC, BD.

Join AD. Then the ext. L AEC=the sum of the int. opp. _8 EDA, EAD;

that is, the sum of the CDA, BAD.

But the 48 CDA, BAD are the angles at the Oce subtended by the arcs AC, BD; :: their sum=half the sum of the angles at the centre subtended by

the same arcs ; or, the L AEC=the angle at the centre subtended by half the sum of the arcs AC, BD.

Q.E.D.

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2. If two chords when produced intersect outside a circle, they form an angle equal to that at the centre subtended by half the difference of the arcs they cut off.

3. The sum of the arcs cut off by two chords of a circle at right angles to one another is equal to the semi-circumference.

4. AB, AC are any two chords of a circle ; and P, Q are the middle points

of the minor arcs cut off by them : if PQ is joined, cutting AB and AC at X, Y, shew that AX=AY.

5. If one side of a quadrilateral inscribed in a circle is produced, the exterior angle is equal to the opposite interior angle.

6. If two circles intersect, and any straight lines are drawn, one through each point of section, terminated by the circumferences ; shew that the chords which join their extremities towards the same parts are parallel.

7. ABCD is a quadrilateral inscribed in a circle; and the opposite sides AB, DC are produced to meet at P, and CB, DA to meet at Q: if the circles circumscribed about the triangles PBC, QAB intersect at R, shew that the points P, R, Q are collinear.

8. If a circle is described on one of the sides of a right-angled triangle, then the tangent drawn to it at the point where it cuts the hypotenuse bisects the other side.

9. Given three points not in the same straight line : shew how to find any number of points on the circle which passes through them, without finding the centre.

10. Through any one of three given points not in the same straight line, draw a tangent to the circle which passes through them, without finding the centre.

11. Of two circles which intersect at A and B, the circumference of one passes through the centre of the other : from A any straight line is drawn to cut the first at C, the second at D; shew that CB=CD.

12. Two tangents AP, AQ are drawn to a circle, and B is the middle point of tħe arc PQ, convex to A. Shew that PB bisects the angle APQ.

13. Two circles intersect at A and B; and at A tangents are drawn, one to each circle, to meet the circumferences at C and D; if CB, BD are joined, shew that the triangles ABC, DBA are equiangular to one another.

14. Two segments of circles are described on the same chord and on the same side of it; the extremities of the common chord are joined to any point on the arc of the exterior segment: shew that the arc intercepted on the interior segment is constant.

H.S.E.

15. If a series of triangles are drawn standing on a fixed base, and having a given vertical angle, show that the bisectors of the vertical angles all pass through a fixed point.

16. ABC is a triangle inscribed in a circle, and E the middle point of the arc subtended by BC on the side remote from A: if through E a diameter ED is drawn, shew that the angle DEA is half the difference of the angles at B and C. [See Ex. 7, p. 109.]

17. If two circles touch each other internally at a point A, any chord of the exterior circle which touches the interior is divided at its point of contact into segments which subtend equal angles at A.

18. If two circles touch one another internally, and a straight line is drawn to cut them, the segments of it intercepted between the circumferences subtend equal angles at the point of contact.

THE ORTHOCENTRE OF A TRIANGLE.

19. The perpendiculars drawn from the vertices of a triangle to the opposite sides are concurrent.

In the A ABC, let AD, BE be the perps drawn from A and B to the opposite sides ; and let them intersect at O. Join CO; and produce it to meet AB at F.

It is required to shew that CF is perp. to AB. Join DE.

B

D C Then, because the 28 OEC, ODC are rt. angles,

Нур. .
:. the points O, E, C, D are concyclic :
:: the L DEC=the L DOC, in the same segment;

=the vert. opp. L FOA. Again, because the 48 AEB, ADB are rt. angles,

Hyp. :. the points A, E, D, B are concyclic :

.: the L DEB=the 2 DAB, in the same segment. is the sum of the 48 FOA, FAO=the sum of the L8 DEC, DEB

=a rt. angle:

Нур. : the remaining - AFO=a rt. angle :

that is, CF is perp. to AB. Hence the three perps AD, BE, CF meet at the point O. Q.E.D.

I. 32.

[For an Alternative Proof see p. 114.]

DEFINITIONS. (i) The intersection of the perpendiculars drawn from the vertices of a triangle to the opposite sides is called its orthocentre.

(ii) The triangle formed by joining the feet of the perpendiculars is called the pedal or orthocentric triangle.

20. In an acute-angled triangle the perpendiculars drawn from the vertices to the opposite sides bisect the angles of the pedal triangle through which they pass.

In the acute-angled ABC, let AD, BE, CF be the perps drawn from the vertices to the opposite sides, meeting at the orthocentre 0; and let DEF be the pedal triangle. Then shall AD, BE, CF bisect respectively the 48 FDE, DEF, EFD.

For, as in the last theorem, it may B be shewn that the points O, D, C, E are concyclic;

:: the L ODE=the L OCE, in the same segment. Similarly the points O, D, B, F are concyclic ;

: the ODF=the L OBF, in the same segment. But the L OCE=the L OBF, each being the compt of the L BAC.

:: the L ODE=the L ODF. Similarly it may be shewn that the 28 DEF, EFD are bisected by BE and CF.

Q.E.D. COROLLARY. (i) Every two sides of the pedal triangle are equally inclined to that side of the original triangle in which they meet. For the L EDC=the compt of the LODE

=the compt of the L OCE

=the 2 BAC. Similarly it may be shewn that the L FDB=the L BAC,

:: the L EDC=the L FDB=the L A. In like manner it may be proved that

the L DÉC=the L FEA=the LB, and the L DFB=the LEFA=the L C.

COROLLARY. (ii) The triangles DEC, AEF, DBF are equiangular to one another and to the triangle ABC.

NOTE. If the angle BAC is obtuse, then the perpendiculars BE, CF bisect externally the corresponding angles of the pedal triangle.

I. 32.

1. 26.

21. In any triangle, if the perpendiculars drawn from the vertices on the opposite sides are produced to meet the circumscribed circle, then each side bisects that portion of the line perpendicular to it which lies between the orthocentre and the circumference.

Let ABC be a triangle in which the perpendiculars AD, BE are drawn, intersecting at O the orthocentre, and let AD be produced to meet the Oce of the circumscribing circle at G.

Then shall DO=DG.
Join BG.

B В
Then in the two As OEA, ODB,
the L OEA=the L ODB, being rt. angles ;
and the L EOA=the vert. opp. - DOB ;
:: the remaining - EÃO=the remaining – DBO.

But the L CAG=the L CBG, in the same segment;
.. the LDBO=the L DBG.
Then in the As DBO, DBG,
the LDBO=the 4 DBG,

Proved.
Because the L BDO=the L BDG,

and BD is common;
:: DO=DG.

Q.E.D. 22. In an acute-angled triangle the three sides are the external bisectors of the angles of the pedal triangle : and in an obtuse-angled triangle the sides containing the obtuse angle are the internal bisectors of the corresponding angles of the pedal triangle.

23. If O is the orthocentre of the triangle ABC, shew that the angles BOC, BAC are supplementary.

24. If O is the orthocentre of the triangle ABC, then any one of the four points O, A, B, C is the orthocentre of the triangle whose vertices are the other three.

25. The three circles which pass through two vertices of a triangle and its orthocentre are each equal to the circle circumscribed about the triangle.

26. D, E are taken on the circumference of a semicircle described on a given straight line AB: the chords AD, BE and AE, BD intersect (produced if necessary) at F and G: shew that FG is perpendicular to AB.

27. ABCD is a parallelogram ; AE and CE are drawn at right angles to AB, and CB respectively: shew that ED, if produced, will be perpendicular to AC.

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