First Method. When The Sides Intersect

Draw the lines 0 C and 0 A - about 3 inches long - forming any angle of 45 to 60 degrees. With 0 as a center and any convenient radius - about 2 inches - draw an arc intersecting the sides of the angle at E and F. With E and F as centers and a radius of 1 1/2 or 1 3/4 inches, describe short arcs intersecting at J. A line 0 D, drawn through the points 0 and I, bisects the angle.

In solving this problem the arc E F should not be too near the vertex if accuracy is desired.

Proof. The central angles A 0 D and DOC are equal because the arc E F is bisected by the line 0 D. The point I is equally distant from E and F.

Second Method. When The Lines Do Not Intersect

Draw the lines A C and E F about 4 inches long making an angle approximately as shown. Draw A' C and E' F' parallel to AC and E F and at such equal distances from them that they will intersect at O. Sow bisect the angle C 0 F' by the method given in Problem 8. The line 0 R bisects the given angle.

Proof. Since A' C" is parallel to A C and E' F' is parallel to E F the angle C O F' is equal to the angle formed by the lines .1 C and E F. Hence as 0 R bisects angle C" O F' it also bisects the angle formed by the lines A C and E F.

Problem 10. To divide a line into any number of equal parts.

Let A C - about 3 3/4 inches long - be a given line. Suppose it is desired to divide it into 7 equal parts. First draw the line A J at least 4 inches long, forming any convenient angle with A C. On A J lav off, by means of the dividers or scale, points D, E, F, G, etc., each \ inch apart. (If dividers are used, the spaces need not be exactly 1/2 inch.) Draw the line J C and through the points D, E, F, G, etc., draw lines parallel to J C. These parallels will divide the line A C into 7 equal parts.

Proof. If a series of parallel lines, cutting two straight lines, intercept equal distances on one of these lines, they also intercept equal distances on the other.

Problem II. To construct a triangle having given the three sides.

Draw the three sides, .1 C, 2 3/4 inches long; E F, 1 15/16 inches long; and M N, 2 3/16 inches long.

Draw R S equal in length to A C. With R as a center and a radius equal to E F describe an arc. With S as a center and a radius equal to M N draw an arc cutting the arc previously drawn, at T. Connect T with R and S to form the triangle.

Problem 12. To construct a triangle having given one side and the two adjacent angles.

Draw the line M N 3 1/4 inches long and draw two angles A 0 D and EFG about 30 degrees and 60 degrees respectively.

Draw R S equal in length to M N and with R as a vertex and R S as one side construct an angle equal to A O D. In a similar manner construe! at S an angle equal to E F G. Draw lines from R and S through the two established points until they meet at T. The triangle R T S will be the required triangle.

Lettering. Draw the pencil lines and put in the lettering as in plates already drawn.

Inking. In inking Plate V, follow the principles previously used and do not make certain lines dotted until sure that they should be dotted.

After inking the figures, ink in the border lines and the lettering as already explained.

Plate VI

Penciling. Lay out this plate in the same manner as the preceding plates.

Problem 13. To describe an arc or circumference through three given points not in the same straight line.

Locate the three points A, B, and C with a distance between A and B of about 2 inches and a distance between A and C of about 2 1/4 inches. Connect A and B and A and C. Erect perpendiculars to the middle points of A B and A C as explained in Problem 1. Now draw light pencil lines connecting the intersections I and J and E and F. These lines will intersect at 0.

With 0 as a center and a radius equal to the distance O A, describe the circumference passing through A, B, and C.

Proof. The point 0 is equally distant from A, B, and C, since it lies in the perpendiculars to the middle points of A B and A C. Hence the circumference will pass through A, B, and C.

Problem 14. To inscribe a circle in a given triangle.

Draw the triangle L M N of any convenient size. M N may be made 3 1/4 inches, L M, 2 3/4 inches, and L N, 3 1/2 inches. Bisect the angles M L N and L M N by the method used in Problem 8. The bisectors M I and L J intersect at 0, which is the center of the inscribed circle. The radius of the circle is equal to the perpendicular distance from 0 to one of the sides.

Proof. The point of intersection of the bisectors of the angles of a triangle is equally distant from the sides.

Problem 15. To inscribe a regular pentagon in a given circle.

With 0 as a center and a radius of about 1 V inches, describe the given circle. With the T-square and triangles draw the center lines A C and E F perpendicular to each other and passing through O. Bisect one of the radii, 0 C, at H and with this point as a center and a radius H E, describe the arc E P. This arc cuts the diameter A C at P. With E as a center and a radius E P, draw arcs cutting the circumference at L and Q. With the same radius and centers at L and Q, draw the arcs cutting the circumference at M and N.

The pentagon is completed by drawing the chords EL, L M, M N, N Q, and Q E.

Problem 16. To inscribe a regular hexagon in a given circle.

With O as a center and a radius of 1 3/8 inches draw the given circle. With the T-square draw the diameter A D. With D as a center, and a radius equal to 0 D, describe arcs cutting the circumference at C and E. Now with C and E as centers and the same radius, draw the arcs, cutting the circumference at B and F. Draw the hexagon by joining the points thus formed.

Therefore, in order to inscribe a regular hexagon in a circle, mark off chords equal in length to the radius.

To inscribe an equilateral triangle in a circle the same method may be used, the triangle being formed by joining the opposite vertices of the hexagon.