the sides, and mark off on one of them a line, A C, equal to one of the

side CB be less than the perpendicular from C on A E, the solution is impossible.

B

E

FIG. 84.

2nd. If the side CB be equal to the perpendicular, the case is possible, and a right-angled triangle only will fulfil the conditions.

3rd. If a be greater than b, the arc cuts AE in one point only, and one triangle can be drawn fulfilling the conditions (Fig. 84).

4th. If a be greater than the perpendicular, but less than b, the arc cuts AE in the points B and B', and either of the triangles, ABC, A B' C, fulfils the given conditions.

To construct a right-angled Triangle, having given the hypothenuse and one angle adjacent to it.

Draw a line equal to the hypothenuse, and make at one extremity an angle equal to the given angle; from the other extremity draw a perpendicular to the opposite side (Fig. 85). It is evident that there is only one solution.

In a triangle there are six parts, namely, three sides and three angles; and we have considered every possible arrangement of three that can be obtained from these six parts, except that of the three angles. It is evident that the three angles of a triangle may be respectively equal to the three angles of another, while the triangles are unequal

(Fig. 86). Hence a triangle is not determined when its three angles are known.

Fractical Application of Problems. In putting these problems into practical use, all lines which have been introduced simply for the purpose of demonstration may be omitted. For example, all the lines required in bisecting an angle are those in Fig. 87.

Again there may be restrictions in practice which make special construction necessary. For instance, let it be required to draw a straight line which shall bisect the angle between two given straight lines which cannot be produced to meet.

FIG. 87.

The bisecting line may be obtained by means of Theorem XXII. Now all points which are at the same distance from a straight line are upon a parallel to this straight line; hence to

find a point in the line bisecting the angle between the two lines A B, C D (Fig. 88), draw a parallel, A'B', to one of the sides AB; at any point D in CD erect a perpendicular, DD, and from this perpendicular cut off a line equal to the distance between A'B' and AB, and finally through D' draw D'C' parallel to DC: the point O where the two parallels intersect, being equidistant from the two sides of the angle, is in the bisecting line. Repeat the operation, making use as much as possible of the construction already made, so as to obtain a second point, Oʻ. O O' will be the bisecting line.

FIG. 88.

The Analytical Method of Solving Problems.-When the steps of

a problem are not at first very evident, it is generally convenient to reverse the direct process of construction thus:

First, sketch a figure, and suppose it to fulfil the given conditions. Secondly, investigate the properties of this figure, with the view to discover the most convenient method of drawing it.

Thirdly, apply the results of the investigation to construct a new figure, step by step, from the given conditions.

For example, take Problem 27 of the following exercise :-« Tɔ construct a triangle, having given the base, the angle opposite the base, and the sum of the sides which contain this angle."

B

A

FIG. 89.

Therefore

D

Suppose A B C to be the triangle required, BC being the given base, BAC the given angle, and B A, AC the sides whose sum is given. Produce B A to D, making A D = A C, and join DC.

Since A D = AC, LADC= LACD.
BAC=▲ADC+ZACD.

But

ADC half ▲ BA C.

We see, therefore, that in ABD C, BC is the given base, ▲ BDC the given angle, and BD the given sum of the two sides.

Construction.--Draw a line, BD, equal to the given sum; at D, make BDC equal to half the given angle; draw B C equal to the given base, by taking the given base as radius, and, with B as centre, describing an arc cutting DC in C. Finally, at C, make DCA=4CDA. Then A ABC shall fulfil the required conditions.

Analysis precedes Construction.-Even when the analysis of a problem is not stated, it always has to be made; for, except by mere trials, no problem could be constructed without previous knowledge of some properties of the figure it is proposed to construct. In every problem a line or figure has to be made which will fulfil given conditions, and the reason for the operations is that they together fulfil the prescribed conditions. If any problem whatever be selected, and at each step of the construction the question be asked, "Why the step is taken ?" it will, in all cases, be found that it is because of some known property of the figure required of which the inventor of the process must have been in possession. Hence it is universally essential to all geometrical construction that the theorems embodying the properties of the figure shall be considered before the actual construction is atten pted.

PROBLEMS FOR EXERCISE.

25. Construct a right-angled triangle, having given one of the sides and the difference of the hypothenuse and the other side.

26. Construct a right-angled triangle, having given the angles and the difference of the hypothenuse and one of the other sides.

27. Construct a triangle, having given the base, the sum of the sides, and the angle opposite the base.

28. Construct a triangle, having given the perimeter, one angle, and a side opposite the angle.

29. Construct a triangle, having given one angle, the perimeter, and the altitude of the triangle from one of the other angles.

30. Construct a triangle, having given the sum of two sides and the angles.

31. Construct a triangle, having given the angles and the perimeter.

CHAPTER IX.

THE CIRCLE.

SECTION I.-Chords and Secants.

Definitions. A straight line drawn within the circle and terminated by the circumference is termed a chord. A straight line drawn right across the circle, so as to cut the circumference in two points, is termed a secant. A chord produced both ways is a secant.

Symmetry of the Circle.-A circle is symmetrical about any diameter --that is to say, if it be folded about any diameter the upper part will exactly fit the lower. From this symmetry we may deduce the following facts respecting chords :

A chord at right angles to a diameter is bisected by the diameter; and, conversely, a line bisecting a chord at right angles is a diameter. Parallel chords intercept equal arcs.

Chords which are at the same distance from the centre are equal. The Diameter the greatest Chord.-The diameter of a circle is its

A

B

FIG. 90.

greatest chord; consequently the diameter is known when we have determined the length of the greatest chord. This fact enables us to measure the diameter when the circle is the section of a rod, a pipe, or a metal wire.

Calibers. For this purpose compasses of a peculiar form are used, called calibers (Fig. 90). Suppose we require to ascertain the diameter or a screw nut, a metal rod, or disc, the compasses are closed in such a manner that the rod may reopen them in passing between the two legs; the distance between the points, measured upon a scale, will give the diameter. In order that the section of the rod may be a true circle, all its diameters must be equal; this may be