Fig. 2.-Let ABC be any angle. With the meeting point or vertex, B, as centre, and any radius, draw an arc cutting the lines of the angle in D and E. Then, as before, with D and E as centres and radius greater than half D E, draw arcs to cut, as shown at F. Join B F. This line will bisect the angles, and any point on it will be equidistant from BD and BE. A repetition of this method will divide the angle into four, or eight equal parts.

(Note that this construction is only applicable when the required number of parts is even, and equal to some integral power of 2, as 22, 23, 24 4, 8, 16, &c.)

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The following points should be carefully observed in connection with the division of lines, arcs, and angles

The exact point at which two lines meet is more accurately found when the angle between them is not less than about 10° or greater than 130°. This will be seen on reference to Fig. 3.

A

B

Fig. 2.

X

Fig. 3.

where the lines A B, CD give a bad meeting point, and the lines E F, G H a good one. It will be seen that, except when the angle is 90°, the lines are in contact for a length greater than their thickness; hence they do not give a decided point of intersection. The same remark applies to the intersection of

arcs.

In bisecting angles it is necessary to obtain the bisecting point (F in Fig. 2) a good distance from the vertex, B, of the angle. If B and F are near together then the line drawn through them will most probably not fulfil the condition that any point in it shall be equidistant from the lines of the angle, except near the vertex. A little practice will soon convince the student of observing this and similar facts. Exactly in the same way, the bisection of a line, by the method just described, is likely to be more accurate when the radius of the arcs is considerably greater than half the line, than when it only slightly exceeds the half, as the former gives a clearly defined intersection point, and the latter a bad one. Other facts of this kind will be referred to in connection with later

problems, but their intelligent study cannot be too soon, or too much, insisted upon.

EXAMPLES.

EX. 1.-Divide a line 3" long into four equal parts by continued bisection.

EX. 2.-Draw an arc of 31′′ radius and bisect it.

EX. 3.-Draw two lines meeting at any angle, and divide the angle into four equal parts by continued bisection.

EX. 4.-Construct a triangle, sides 5′′, 4′′, and 3′′, and bisect each of its angles. The bisecting lines will meet at a point. Show, by drawing the circle, that this point is the centre of the inscribed circle of the triangle.

(Obtain the radius of the circle by drawing a perpendicular with set squares from its centre to one of the sides.)

EX. 5.-Draw an arc of 41" radius, and divide it into four equal parts by continued bisection. Produce the bisecting lines and notice that they pass through the centre of the arc.

Perpendiculars to lines-PROBLEM II. (Figs. 4a, b, c, d).To draw a perpendicular to a given line, A B, from a given point, C. (a) (Fig. 4a).—When the point is in the line and not near either end.

With the given point C as centre, and any radius less than the shorter of C A or CB, cut the line A B in the points D and E, on

opposite sides of C. With D and E as centre, and radius greater than half D E, draw arcs meeting on either side of the line, as shown in F. The line through FC is perpendicular to A B.

(b) (Fig. 4b).—When the point is outside the line and not over either end.

With the given point C as centre, describe an arc cutting the line in two points, D and E. With these points as centres, and

with the same radius, draw arcs on the other side of A B, meeting in F. The line through C F is perpendicular to A B.

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(Euclid i. 8, and Def. 10, for CD CE, and FD FE, and CF is common, therefore the angles made by A B and C F are right angles). (c) (Fig. 4c). When the point is in the line, and near to, or at one end.

With the given point C as centre, and any radius less than C A, draw an arc as shown, cutting C A in D. From D step off the same radius from D to E, and E to F. With E and F as centre, and any radius (for convenience the same as before), draw arcs cutting at G. The line through CG is perpendicular to A B. (The angle DCE=60°, also the angle ECF. But CG bisects the angle ECF making angle ACG 60° + 30° 90°).

(d) Fig. 4d). When the point is outside the line and over either

Join the given point C to any convenient point D, near the further end of AB. Bisect CD in E, and with E as centre draw a semicircle passing through C and D, and cutting A B in F. The line through CF is perpendicular to A B.

(The angle in a semicircle is a right angle (Euclid iii., 31), and as C F D is an angle in a semicircle, C F is perpendicular to A B).

This construction can also be applied to Case C, Fig. 4c, by drawing any semicircle with centre E passing through the given point C, and cutting the line A B in a point D. Then by joining the points DE, and producing the line to cut the semicircle at a second point, G, a diameter will be drawn, and G will be the required point to join to C.

Notice that what has already been said about choosing the radii of the arcs so as to obtain sharp points of intersection applies equally to these problems, and that it is only necessary to draw part of the arcs through where the cutting point is likely to come. Also that in such an example as Fig. 4b, time

is saved by using the same radius for the arcs, cutting at Fas for the arc described from centre C, thus avoiding an alteration of the compasses.

EXAMPLES.

The following Examples are to be constructed geometrically:-EX. 6. Construct a square of 21′′ side.

EX. 7.--Construct an oblong, sides 3′′ and 2′′.

EX. 8.-Construct a triangle, sides 5′′, 4′′, 3′′, and draw from each corner a perpendicular to the opposite side.

EX. 9.-Construct a triangle, sides 5′′, 44′′, 23′′. Bisect each of its sides, and through the points draw perpendiculars to the sides. These three lines will meet at a point. Show, by drawing the circle, that this point is the centre of the circle circumscribing the triangle.

EX. 10.-Mark any three points A, B, and C, not in the same straight line, and draw the circle passing through them.

(Find the centre by joining A to B and B to C, and bisect these lines by lines perpendicular to them; they will meet at the centre.)

EX. 11.-Draw any irregular triangle, and mark any point within it. From this point draw lines perpendicular to each of the sides.

Parallels to Lines-PROBLEM III. (Fig. 5).-To draw a line parallel to the given line A B at a given distance away.

Construct perpendiculars, AD,

BC, from the ends of the line, or from any convenient points within it, and cut off a length on each equal to the given distance. Then through the two points draw a line which will be parallel to A B;

or

Fig. 5.

Draw a perpendicular from one end of the line as B C, cut off a length equal to the given distance, as B E, and through point E draw another perpendicular to BC, which will be parallel to AB. As both these methods require the construction of two perpendiculars, there is no reason why they should not be equally

accurate.

EXAMPLE.

Then

EX. 12-Draw a rhombus, sides 31", acute angles 45°. draw a second rhombus, parallel to and surrounding it, and " away.

(Obtain the 45° by bisecting a right angle.)

Copying and Addition of Angles.-Similar straight-lined

figures are those which have their several angles equal, each to each, and the sides about the equal angles proportional to each other. Hence to draw one figure similar to another, it is necessary to know how to copy an angle, or, in other words, how to make an angle at a given point equal to a given angle. It is also convenient to be able, geometrically, to add angles together, as, for example, an angle of 135° can be found by adding angles of 90° and 45°; and 75°, by adding 60° and the fourth part of 60°. PROBLEM IV. (Fig. 6).-On a given line to draw a triangle having angles equal to those of a given triangle.

Let A B C be the given triangle and D E the given line.

With centre B, draw an arc cutting BA, BC, in F and G. Draw a similar arc with the same radius, with centre D, cutting DE in H. Measure the chord, FG, in the compasses, and with H as centre set off the length of F G along the arc to K. Join D K and produce. Then the angle KDE is equal to the angle ABC. At E, make an angle equal to the angle at C or A in the same way, and the triangle will be complete. (Euclid vi., 4.) (Equal angles in equal circles are subtended by equal chords.-Euclid iii., 26 and 29.)

Note. If the triangles are large, accuracy will only be obtained by drawing arcs of large radius.

PROBLEM V. (Fig. 7).-To add or subtract angles.

H

Fig. 7.

M

Let A and B be two angles. It is required to make with the line CD an angle equal to A and B.

With A, B, and C as centres, draw arcs of the same radius, KID cutting the arms of A in E F and the arms of B in G H, and cutting the line CD in K. Take the length of the arc,

EF, in the compasses and set it up from K to L.

Similarly,