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Because SH and OU are both 1 BC .. SH is || OU;

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SK

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CA SKOV. Hence the AS SHK, OUV are mutually equiangular, and since HK = UV. SH = OU

.'.

= half AO.

I. 28, Cor.

I. 28, Cor.

I. 34, Cor.

I. 26

Again, because P and Q are the middle points of GA, GO;

PQ is | AO and half AO;

.. PQ is || SH and

=

= SH.

Hence the As HGS, PGQ are equal in all respects;

.. HG PG half AG;

=

=

.. G is the centroid,

App. I. 1

I. 29, 26

App. I. 4

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COR.-The distance of the circumscribed centre from any side of a triangle is half the distance of the orthocentre from the opposite vertex.

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Many of the problems which occur in geometry consist in the finding of points. Now the position of a point-and position is the only property which a point possesses-is determined by certain conditions, and if we know these conditions, we can, in general, find the point which satisfies them. It will be seen that in plane geometry two conditions suffice to determine a point, provided the conditions be mutually consistent and independent. When only one of the conditions is given, though the point cannot then be determined, yet its position may be so restricted as to enable us to say that wherever the point may be, it must always lie on some one or two lines which we can describe; for example, straight lines

or the circumferences of circles. The given condition may, however, be such that the point which satisfies it will lie on a line or lines which we do not as yet know how to describe. Cases where this occurs are considered as not belonging to elementary plane geometry.

DEF. The line (or lines) to which a point fulfilling a given condition is restricted, that is, on which alone it can lie, is (or are) called the locus of the point. Instead of the phrase 'the locus of a point,' we frequently say 'the locus of points.'

For the complete establishment of a locus, it ought to be proved not only that all the points which are said to constitute the locus fulfil the given condition, but that no other points fulfil it. The latter part of the proof is generally omitted.

Ex. 1. Find the locus of a point having the property (or fulfilling the condition) of being situated at a given distance from a given point. Let A be the given point, and suppose

B, C, D, &c. to be points on the locus. Join

AB, AC, AD, &c.

Then AB = AC AD = &c. ;

=

Hyp.

and hence B, C, D, &c. must be situated on the Oce of a circle whose centre is A, and whose radius is the given distance.

AB, AC, AD, &c.

Moreover, the distance from A of any point not situated on the Oce would not be = This Oce.. is the required locus.

D

C

B

Ex. 2. Find the locus of a point having the property (or fulfilling the condition) of being equidistant from

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Hence, if a set of other points on the locus be taken, and joined to the definite fixed point C, a set of perpendiculars to AB will be obtained. The locus therefore consists of all the perpendiculars that can be drawn to AB through the point C; that is, CD produced indefinitely either way is the locus.

PROPOSITION 6.

Straight lines are drawn from a given fixed point to the circumference of a given fixed circle, and are bisected: find the locus of their middle points.

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Let A be the given fixed point, C the centre of the given fixed circle; let AB, one of the straight lines drawn from A to the Oce, be bisected at E:

it is required to find the locus of E.

Join AC, and bisect it at D;

I. 10

join DE and CB.

Because DE joins the middle points of two sides of ▲ ACB,

... DE =

CB.

App. I. 1

But CB, being the radius of a fixed circle, is a fixed length;

.. DE, its half, is also a fixed length.

Again, since A and C are fixed points,

.. AC is a fixed straight line;

.. D, the middle point of AC, is a fixed point;

that is, E, the middle point of AB, is situated at a fixed distance from the fixed point D.

But AB was any straight line drawn from A to the Oce;

.. the middle points of all other straight lines drawn from A to the Oce must be situated at the same fixed distance from the fixed point Ꭰ ;

.. the locus of the middle points is the Oce of a circle, whose centre is D, and whose radius is half the radius of the fixed circle. From the figure it will be seen that it is immaterial whether AB or AB' is to be considered as the straight line drawn from A to the Oce. For if E be the middle point of AB', then E'D = B'C,

that is half the radius of the fixed circle;

=

.. the locus of E' is the same Oce as before.

[The reader is requested to make figures for the cases when the given point A is inside the given circle, and when it is on the Oce of the given circle.]

INTERSECTION OF LOCI.

Since two conditions determine a point, if we can construct the locus satisfying each condition, the point or points of intersection of the two loci will be the point or points required. A familiar example of this method of determining a point, is the finding of the position of a town on a map by means of parallels of latitude and meridians of longitude. The reader is recommended to apply this method to the solution of I. 1 and 22, and to several of the problems on the construction of triangles.

DEDUCTIONS.

1. The straight line joining the middle points of the non-parallel sides of a trapezium is || the parallel sides and = half their

sum.

=

2. The straight line joining the middle points of the diagonals of a trapezium is the parallel sides and half their difference. 3. The straight line joining the middle points of the non-parallel sides of a trapezium bisects the two diagonals.

4. The middle points of any two opposite sides of a quadrilateral and the middle points of the two diagonals are the vertices of a m

5. The straight lines which join the middle points of the opposite sides of a quadrilateral, and the straight line which joins the middle points of the diagonals, are concurrent.

=

6. If from the three vertices and the centroid of a triangle perpendiculars be drawn to a straight line outside the triangle, the perpendicular from the centroid one-third of the sum of the other perpendiculars. Examine the cases when the straight line cuts the triangle, and when it passes through the centroid.

7. Find a point in a given straight line such that the sum of its distances from two given points may be the least possible. Examine the two cases, when the two given points are on the same side of the given line, and when they are on different sides.

8. Find a point in a given straight line such that the difference of its distances from two given points may be the least possible. Examine the two cases.

9. Of all triangles having only two sides given, that is the greatest in which these sides are perpendicular.

10. The perimeter of an isosceles triangle is less than that of any other triangle of equal area standing on the same base.

11. Of all triangles having the same vertical angle, and the bases of which pass through the same given point, the least is that which has its base bisected by the given point.

12. Of all triangles formed with a given angle which is contained by two sides whose sum is constant, the isosceles triangle has the least perimeter.

13. The sum of the perpendiculars drawn from any point in the base of an isosceles triangle to the other two sides is constant. Examine the case when the point is in the base produced.

14. The sum of the perpendiculars drawn from any point inside an equilateral triangle to the three sides is constant. Examine the case when the point is outside the triangle.

15. The sum of the perpendiculars from the vertices of a triangle on the opposite sides is greater than the semi-perimeter and less than the perimeter of the triangle.

16. If a perpendicular be drawn from the vertical angle of a triangle to the base, it will divide the vertical angle and the base into parts such that the greater is next the greater side of the triangle.

17. The bisector of the vertical angle of a triangle divides the base into segments such that the greater is next the greater side of the triangle.

18. The median from the vertical angle of a triangle divides the vertical angle into parts such that the greater is next the less side of the triangle.

19. If from the vertex of a triangle there be drawn a perpendicular to the opposite side, a bisector of the vertical angle and a median, the second of these lies in position and magnitude between the other two.

20. The sum of the three angular bisectors of a triangle is greater than the semiperimeter, and less than the perimeter of the triangle.

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