Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

36. Bisect a triangle by a straight line drawn through a given point in one of its sides.

[Let ABC be the given A, and P the
given point in the side AB.

Bisect AB at Z; and join CZ, CP.
Through Z draw ZQ parallel to CP.

Join PQ.
Then shall PQ bisect the A.

See Ex. 21, p. 119.] B

[ocr errors]

37. Trisect a triangle by straight lines drawn from a given point in one of its sides.

[Let ABC be the given A, and X the
given point in the side BC.
Trisect BC at the points P, Q. Ex. 19, p. 107.
Join AX, and through P and Q draw PH

Н.

K and QK parallel to AX.

Join XH, XK.
These straight lines shall trisect the A ; as
may be shewn by joining AP, AQ.

B
See Ex. 21, p. 119.]

38. Cut off from a given triangle a fourth, fifth, sixth, or any part required by a straight line drawn from a given point in one of its sides.

[See Ex. 19, p. 107, and Ex. 21, p. 119.]

39. Bisect a quadrilateral by a straight line drawn through an angular point.

[Two constructions may be given for this problem : the first will be suggested by Exercises 28 and 33, p. 120.

The second method proceeds thus.

Let ABCD be the given quadri.
lateral, and A the given angular point.

D
Join AC, BD, and bisect BD in X.
Through X draw PXQ parallel to AC,

meeting BC in P; join AP.
Then shall AP bisect the quadrilateral.
Join AX, CX, and use i. 37, 38.]

A

40. Cut off from a given quadrilateral a third, a fourth, a fifth, or any part required, by a straight line drawn through a given angular point.

(See Exercises 28 and 35, p. 120.]

Obs. The following Theorems depend on I. 47. V 41. In the figure of 1. 47, shew that (i) the sum of the squares on AB and AE is equal to the sum

of the squares on AC and AD. (ii) the square on EK is equal to the square on AB with four

times the square on AC. (iii) the sum of the squares on EK and FD is equal to five

times the square on BC.

42. If a straight line is divided into any two parts, the square on the straight line is greater than the sum of the squares on the

two parts. V 43. If the square on one side of a triangle is less than the

squares on the remaining sides, the angle contained by these sides is

acute; if greater, obtuse. v 44. ABC is a triangle, right-angled at A; the sides AB, AC are

intersected by a straight line PQ, and BQ, PC are joined : shew that the sum of the squares on BQ, PC is equal to the sum of the squares on BC, PQ.

45. In a right-angled triangle four times the sum of the squares on the medians which bisect the sides containing the right angle is equal to five times the square on the hypotenuse.

46. Describe a square whose area shall be three times that of a given square.

47. Divide a straight line into two parts such that the sum of their squares shall be equal to a given square.

a

[blocks in formation]

In many geometrical problems we are required to find the position of a point which satisfies given conditions ; and all such problems hitherto considered have been found to admit of a limited number of solutions. This, however, will not be the case if only one condition is given. For example:

(i) Required a point which shall be at a given distance from a given point.

This problem is evidently indeterminate, that is to say, it admits of an indefinite number of solutions; for the condition stated is satisfied by any point on the circumference of the circle described from the given point as centre, with a radius equal to the given distance. Moreover this condition is satisfied by no other point within or without the circle.

(ii) Required a point which shall be at a given distance from a given straight line.

Here again there are an infinite number of such points, and they lie on two parallel straight lines drawn on either side of the given straight line at the given distance from it: further, no point that is not on one or other of these parallels satisfies the given condition.

Hence we see that one condition is not sufficient to determine the position of a point absolutely, but it may have the effect of restricting it to some definite line or lines, straight or curved. This leads us to the following definition.

DEFINITION. The Locus of a point satisfying an assigned con. dition consists of the line, lines, or part of a line, to which the point is thereby restricted; provided that the condition is satisfied by every point on such line or lines, and by no other.

A locus is sometimes defined as the path traced out by a point which moves in accordance with an assigned law.

Thus the locus of a point, which is always at a given distance from a given point, is a circle of which the given point is the centre: and the locus of a point, which is always at a given distance from a given straight line, is a pair of parallel straight lines.

We now see that in order to infer that a certain line, or system of lines, is the locus of a point under a given condition, it is necessary

a

to prove

(i) that any point which fulfils the given condition is on the supposed locus ;

(ii) that every point on the supposed locus satisfies the given condition.

1. Find the locus of a point which is always equidistant from two given points.

Let A, B be the two given points.

(a) Let P be any point equidistant from A and B, so that AP=BP.

Bisect AB at X, and join PX.
Then in the As AXP, BXP,
AX=BX,
Constr. AF

B Because and PX is common to both, also AP=BP,

Нур.
:. the PXA=the PXB;
and they are adjacent 48 ;

:: PX is perp. to AB. Def. 10.
.. any point which is equidistant from A and B
is on the straight line which bisects AB at right angles.

X

1. 8.

(B) Also every point in this line is equidistant from A and B.

For let Q be any point in this line.

Join AQ, BQ.
Then in the A8 AXQ, BXQ,

AX=BX,
Because

and XQ is common to both;
also the L AXQ=the LBXQ, being rt. 28 ;

:: AQ=BQ. That is, Q is equidistant from A and B. Hence we conclude that the locus of the point equidistant from two given points A, B is the straight line which bisects AB at right angles.

1. 4.

2. To find the locus of the middle point of a straight line drawn from a given point to meet a given straight line of unlimited length.

[blocks in formation]

Let A be the given point, and BC the given straight line of unlimited length.

(a) Let AX be any straight line drawn through A to meet BC, and let P be its middle point.

Draw AF perp. to BC, and bisect AF at E.

Join EP, and produce it indefinitely. Since AFX is a A, and E, P the middle points of the two sides AF, AX,

:: EP is parallel to the remaining side FX. Ex. 2, p. 104. :. P is on the straight line which passes through the fixed point E, and is parallel to BC.

(B). Again, every point in EP, or EP produced, fulfils the required condition.

For, in this straight line take any point Q.

Join AQ, and produce it to meet BC in Y. Then FAY is a A, and through E, the middle point of the side AF, EQ is drawn parallel to the side FY;

:: Q is the middle point of AY. Ex. 1, p. 104. Hence the required locus is the straight line drawn parallel to BC, and passing through E, the middle point of the perp. from A to BC. 3. Find the locus of a point equidistant from two given intersecting straight lines.

[See Ex. 3, p. 55.] 4. Find the locus of a point at a given radial distance from the circumference of a given circle.

5. Find the locus of a point which moves so that the sum of its distances from two given intersecting straight lines of unlimited length is constant.

6. Find the locus of a point when the differences of its distances from two given intersecting straight lines of unlimited length is constant.

[ocr errors]

7. A straight rod of given length slides between two straight rulers placed at right angles to one another : find the locus of its middle point.

[See Ex. 2, p. 108.] 8. On a given base as hypotenuse right-angled triangles are described : find the locus of their vertices. [See Ex. 2, p. 108.]

[ocr errors]

a

9. AB is a given straight line, and AX is the perpendicular drawn from A to any straight line passing through B : find the locus of the middle point of AX.

10. Find the locus of the vertex of a triangle, when the base and area are given.

11. Find the locus of the intersection of the diagonals of a parallelogram, of which the base and area are given. 12. Find the locus of the intersection of the medians of triangles described on a given base and of given area.

[ocr errors][merged small][merged small]

It appears from various problems which have already been considered, that we are often required to find a point, the position of which is subject to two given conditions. The method of loci is very useful in solving problems of this kind ; for corresponding to each condition there will be a locus on which the required point must lie. Hence all points which are common to these two loci, that is, all the points of intersection of the loci, will satisfy both the given conditions.

« ΠροηγούμενηΣυνέχεια »