ON THE NINE-POINTS CIRCLE.

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 AABC, 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, D, E, F, a, ß, y be concyclic.

Join XY, XZ, Xa, Ya, Za. Now from the ▲ ABO, since AZ = ZB, and Aa=aO,

Hyp.

. Za is par to BO. Ex. 2, p. 96. And from the A ABC, since BZ = ZA, and BX=XC,

Δ

.. ZX is par1 to AC.

But BO makes a rt. angle with AC;
.. the XZa is a rt. angle.

Similarly, the XYa is a rt. angle.

.. the points X, Z, a, Y are concyclic:

Y

E

γ

Нур.

I. 29.

that is, a lies on the " of the circle, which passes through X, Y, Z ; and Xa is a diameter of this circle.

Similarly it may be shewn that ẞ and y lie on the Oce of the circle which passes through X, Y, Z.

Again, since aDX is a rt. angle,

Hyp.

.. 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, ß, 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 ▲ 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. to XD from its middle point bisects so; Ex. 14, p. 98. Similarly the perp. to EY at its middle point bisects SO:

Y

SK

E

that is, these perp intersect at the middle point of SO: And since XD and EY are chords of the nine-points circle,

O

.. the intersection of the lines which bisect XD and EY at rt. angles is its centre:

.. the centre N is the middle point of SO.

III. 1.

(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.

.. the radius of the nine-points circle is half the radius of the circumscribed circle,

(iii) To prove that the centroid is collinear with points S, N, O. Join AX and draw ag par1 to SO.

Let AX meet SO at G.

Then from the A AGO, since AaaO and ag is par1 to OG,

.. Ag=gG.

Ex. 13, p. 98.

And from the ▲ Xag, since aN NX, and NG is par1 to ag,

.. gG=GX.

.. AG of AX;

.. G is the centroid of the triangle ABC.

Ex. 13, p. 98.

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 1, 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, 13.

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 Ex. 60, p. 382.

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 circumscribed 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, I, I, 12, I are the centres of the inscribed and escribed circles, and S1, S., S, are the centres of the circles circumscribed about the triangles BIC, CIA, AIB: shew that the triangle SSS, has its sides parallel to those of the triangle Ills, and is one-fourth of it in area: also that the triangles ABC and SSS have the same circumscribed circle.

9. O is the orthocentre of a triangle ABC: shew that AO2+ BC2 BO2 + CA2 = CO2 + AB2= d2,

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 PA+ PB+PC is constant,

BOOK V.

Book V. treats of Ratio and Proportion.

INTRODUCTORY.

The first four books of Euclid deal with the absolute equality or inequality of Geometrical magnitudes. In the Fifth Book magnitudes are compared by considering their ratio, or relative greatness.

The meaning of the words ratio and proportion in their simplest arithmetical sense, as contained in the following definitions, is probably familiar to the student:

The ratio of one number to another is the multiple, part, or parts that the first number is of the second; and it may therefore be measured by the fraction of which the first number is the numerator and the second the denominator.

Four numbers are in proportion when the ratio of the first to the second is equal to that of the third to the fourth.

But it will be seen that these definitions are inapplicable to Geometrical magnitudes for the following reasons:

(1) Pure Geometry deals only with concrete magnitudes, represented by diagrams, but not referred to any common unit in terms of which they are measured in other words, it makes no use of number for the purpose of comparison between different magnitudes.

(2) It commonly happens that Geometrical magnitudes of the same kind are incommensurable, that is, they are such that it is impossible to express them exactly in terms of some common unit.

For example, we can make comparison between the side and diagonal of a square, and we may form an idea of their relative greatness, but it can be shewn that it is impossible to divide either of them into equal parts of which the other contains an exact number. And as the magnitudes we meet with in Geometry are more often incommensurable than not, it is clear that it would not always be possible to exactly represent such magnitudes by numbers, even if reference to a common unit were not foreign to the principles of Euclid.

It is therefore necessary to establish the Geometrical Theory of Proportion on a basis quite independent of Arithmetical principles. This is the aim of Euclid's Fifth Book.