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22. On a base of given length describe a triangle equal to a given triangle and having an angle equal to an angle of the given triangle.

23. Construct a triangle equal in area to a given triangle, and having a given altitude.

24. On a base of given length construct a triangle equal to a given triangle, and having its vertex on a given straight line.

25. On a base of given length describe (i) an isosceles triangle ; (ii) a right-angled triangle, equal to a given triangle.

26. Construct a triangle equal to the sum or difference of two given triangles. [See Ex. 16, p. 118.]

27. ABC is a given triangle, and X a given point: describe a triangle equal to ABC, having its vertex at X, and its base in the same straight line as BC.

28. ABCD is a quadrilateral. On the base AB construct a triangle equal in area to ABCD, and having the angle at A common with the quadrilateral.

[Join BD. Through C draw CX parallel to BD, meeting AD produced in X; join BX.]

29. Construct a rectilineal figure equal to a given rectilineal figure, and having fewer sides by one than the given figure.

Hence shew how to construct a triangle equal to a given rectilineal figure.

30. ABCD is a quadrilateral: it is required to construct a triangle equal in area to ABCD, having its vertex at a given point X in DC, and its base in the same straight line as AB.

31. Construct a rhombus equal to a given parallelogram.

32. Construct a parallelogram which shall have the same area and perimeter as a given triangle.

33. Bisect a triangle by a straight line drawn through one of its angular points.

34. Trisect a triangle by straight lines drawn through one of its angular points.

[See Ex. 19, p. 110, and 1. 38.] 35. Divide a triangle into any number of equal parts by straight lines drawn through one of its angular points.

[See Ex. 19, p. 107, and 1. 38.] 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

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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.

K Join AX, and through P and Q draw PH and QK parallel to AX.

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

B Р X
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.

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

meeting BC in P; join AP. Then shail AP bisect the quadrilateral. Join AX, CX, and use 1. 37, 38.]

A

B

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 1. 47. 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.

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. 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.

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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 hy 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 condition 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 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. AS

B

Х
Because and PX common to both,
also AP=BP,

Нур. .
.. the L PXA=the Ź PXB;
and they are adjacent (8 ;

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.

I. 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 As AXQ, BXQ,

AX=BX,
Because

and XQ is common to both;
also the L AXQ=the L BXQ, 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.

I. 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.

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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 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.

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