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ON THE NINE-POINTS CIRCLE.

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32. In any triangle the middle points of the sides, the feet of the perpendiculars drawn from the vertices to the opposite sides, and the middle points of the lines joining the orthocentre to the vertices are concyclic.

In the A ABC, let X, Y, Z be the middle points of the sides BC, CA, AB; let D, E, F be the feet of the perps drawn to these sides from A, B, C; let O be the orthocentre, and a, b, y the middle points of OA, OB, OC. Then shall the nine points X, Y, Z,

E D, E, F, a, b, y be concyclic. Join XY, XZ, Xa, Ya, Za. Now from the A ABO,

B

х since AZ=ZB, and Aa=aO, Hyp. .. Za is par to BO. Ex. 2, p. 104. And from the ABC, since BZ=ZA, and BX=XC, Hyp.

:: ZX is part to AC. But BO produced makes a rt. angle with AC; Hyp.

:: the L XZa is a rt. angle. Similarly, the _ XYa is a rt. angle.

1. 29. .. the points X, Z, a, Y are concyclic: that is, a lies on the Oce of the circle, which passes through X, Y, Z; and Xã is a diameter of this circle.

Similarly it may be shewn that B and y lie on the Oce of the circle which passes through X, Y, Z. Again, since aDX is a rt. angle,

Нур. . :: the circle on Xa as diameter passes through D. Similarly it may be shewn that E and F lie on the circumference of the same circle.

:: the points X, Y, Z, D, E, F, a, b, y are concyclic. Q.E.D.

From this property the circle which passes through the middle points of the sides of a triangle is called the Nine-Points Circle ; many of its properties may be derived from the fact of its being the circle circumscribed about the pedal triangle.

33. To prove that

(i) the centre of the nine-points circle is the middle point of the straight line which joins the orthocentre to the circumscribed centre.

(ii) the radius of the nine-points circle is half the radius of the circumscribed circle.

(iii) the centroid is collinear with the circumscribed centre, the nine-points centre, and the orthocentre.

In the A ABC, let X, Y, Z be the middle points of the sides ; D, E, F the feet of the perps ; O the orthocentre; S and N the centres of the circumscribed and nine-points circles respectively.

(i) To prove that N is the middle point of so. It may be shewn that the perp.

E to XD from its middle point bisects

Ex. 14, p. 106. Similarly the perp. to EY at its B

Х middle point bisects SO:

that is, these perps intersect at the middle point of SO: And since XD and EY are chords of the nine-points circle, .. the intersection of the lines which bisect XD and EY at rt. angles is its centre:

III. 1. :: the centre N is the middle point of SO.

SO;

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(ii) To prove that the radius of the nine-points circle is half the radius of the circumscribed circle. By the last Proposition, Xa is a diameter of the nine-points circle.

:: the middle point of Xa is its centre: but the middle point of SO is also the centre of the nine-points circle.

(Proved.)
Hence Xa and SO bisect one another at N.
Then from the As SNX, ONa,

SN=ON,
Because

and NX=Na,
and the L SNX=the L ONa;
:: SX=Oa

I. 4.
= Aa.
And SX is also parl to Aa,
:. SA=Xa.

I. 33. But SA is a radius of the circumscribed circle ;

and Xa is a diameter of the nine-points circle ; :: the radius of the nine-points circle is half the radius of the cir.

cumscribed circle.

1. 15.

(iii) To prove that the centroid is collinear with points S, N, O.

Join AX and draw ag parl to SO.

Let AX meet so at G.
Then from the A AGO, since Aa=a0, and ag is par to OG,

:: Ag=gG.

Ex. 1, p. 104. And from the A Xag, since aŇ=NX, and NG is parl to ag,

:: gG=GX.

ÄG=ş of AX; :: G is the centroid of the triangle ABC. Ex. 4, p. 113. That is, the centroid is collinear with the points S, N, O. Q.E.D.

34. Given the base and vertical angle of a triangle, find the locus of the centre of the nine-points circle.

35. The nine-points circle of any triangle ABC, whose orthocentre is O, is also the nine-points circle of each of the triangles AOB, BOC, COA.

36. If I, 11, 12, 13 are the centres of the inscribed and escribed circles of a triangle ABC, then the circle circumscribed about ABC is the nine-points circle of each of the four triangles formed by joining three of the points I, 11, 12, 1.3.

37. All triangles which have the same orthocentre and the same circumscribed circle, have also the same nine-points circle.

38. Given the base and vertical angle of a triangle, shew that one angle and one side of the pedal triangle are constant.

39. Given the base and vertical angle of a triangle, find the locus of the centre of the circle which passes through the three escribed centres.

NOTE. For another important property of the Nine-points Circle see Miscellaneous Examples on Book VI., Ex. 60.

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1. If four circles are described to touch every three sides of a quadrilateral, shew that their centres are concyclic.

2. If the straight lines which bisect the angles of a rectilineal figure are concurrent, a circle may be inscribed in the figure.

3. Within a given circle describe three equal circles touching one another and the given circle.

4. The perpendiculars drawn from the centres of the three escribed circles of a triangle to the sides which they touch, are concurrent.

5. Given an angle and the radii of the inscribed and circum. scribed circles ; construct the triangle.

6. Given the base, an angle at the base, and the distance between the centre of the inscribed circle and the centre of the escribed circle which touches the base ; construct the triangle.

7. In a given circle inscribe a triangle such that two of its sides may pass through two given points, and the third side be of given length.

8. In any triangle ABC, 1, 11, 12, 13 are he centres of the inscribed and escribed circles, and S1, S2, Sz are the centres of the circles circumscribed about the triangles BIC, CIA, AIB : shew that the triangle S, S, Sz has its sides parallel to those of the triangle l1/2/3, and is one-fourth of it in area : also that the triangles ABC and S,S,S, have the same circumscribed circle. 9. O is the orthocentre of a triangle ABC: shew that

AO2+ BC =BO2+ CAP=CO2+ AB2=d, where d is the diameter of the circumscribed circle.

10. If from any point within a regular polygon of n sides perpendiculars are drawn to the sides, the sum of the perpendiculars is equal to n times the radius of the inscribed circle.

11. The sum of the perpendiculars drawn from the vertices of a regular polygon of n sides on any straight line is equal to n times the perpendicular drawn from the centre of the inscribed circle.

12. The area of a cyclic quadrilateral is independent of the order in which the sides are placed in the circle.

13. Given the orthocentre, the centre of the nine-points circle, and the middle point of the base; construct the triangle.

14. Of all polygons of a given number of sides, which may be inscribed in a given circle, that which is regular has the maximum area and the maximum perimeter.

15. Of all polygons of a given number of sides circumscribed about a given circle, that which is regular has the minimum area and the minimum perimeter.

16. Given the vertical angle of a triangle in position and magnitude, and the sum of the sides containing it: find the locus of the centre of the circumscribed circle.

17. P is any point on the circumference of a circle circumscribed about an equilateral triangle ABC: shew that PA2+ PB2+ PC2 is constant.

APPENDIX.

ON POLE AND POLAR.

DEFINITIONS.

(i) If in any straight line drawn from the centre of a circle two points are taken such that the rectangle contained by their distances from the centre is equal to the square on the radius, each point is said to be the inverse of the other.

Thus in the figure given below, if O is the centre of the circle, and if OP. OQ=(radius)?, then each of the points P and Q is the inverse of the other.

It is clear that if one of these points is within the circle the other must be without it.

(ii) The polar of a given point with respect to a given circle is the straight line drawn through the inverse of the given point at right angles to the line which joins the given point to the centre: and with reference to the polar the given point is called the pole. Thus in the adjoining figure, if OP. OQ=(radius)?, and if through

L
pl

M

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P and Q, LM and HK are drawn perp. to OP; then HK is the polar of the point P, and P is the pole of the st. line HK: also LM is the polar of the point Q, and Q the pole of LM.

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