1. TO BISECT ANY LINE PERPENDICULARLY

In Figure 19 we see the line A-B. It is required to draw a line through A-B at right angles to it, cutting it in half.

METHOD: With B as a center and any radius greater than half A-B, draw an arc above and below the line A-B. With A as a center and the same radius, do the same thing. The two arcs will meet above and below A-B in the points O and 0'. Now join O with O'; this line bisects A-B at right angles.

2. TO CONSTRUCT AN ANGLE EQUAL TO A GIVEN ANGLE

In Figure 20 we see a given angle XOC and a line X'Y to the right of it. It is required to construct the angle XOC on the line X'Y. METHOD : With O as a center draw any arc of any radius, cutting the line OX in A and the line OC in B. Lay this same arc off on line X'Y and call the pivot point 0'. Now, with B as a center, make an arc equal to BA, and lay this arc off on the arc you have already drawn on X'Y. Now connect O' and A'; the resulting angle equals angle XOC.

3. TO DIVIDE ANY LINE INTO ANY NUMBER OF EQUAL PARTS

In Figure 21 we have a line AB which is 3 13/16 inches long. It is required to divide this line into 17 equal parts.

METHOD: From A draw A-X at any angle at all, and on A-X measure off 17 equal divisions as shown. Now join the 17th division with the point B; through each division draw lines parallel to this end line, making sure that they all cut the line A-B. These lines will divide A-B into 17 equal parts.

4. TO DIVIDE A CIRCLE INTO ANY NUMBER OF EQUAL PARTS

In Figure 22 we have a circle with a diameter A-B. It is required to divide this circle into 11 equal parts.

METHOD: With B as a center, swing a small arc equal to AB below the circle. Do the same with A as a center, and the two arcs will meet at point O. Now divide the diameter A-B into 11 equal parts by the method described in Number 3, above, and call the point where the second mark meets the diameter point E. Draw a line through O and E which will cut the circumference at D. Now join A with D. A-D is one-eleventh of the circumference, and so with your small dividers you can lay off the other points.

Obviously, in 3 and 4 the line A-X can be divided into as many equal divisions as you wish; consequently, the line A-B can also be divided into as many equal divisions as you wish. In the case of the circle you need only consider the second division because that is the one through which the line from O must pass.

5. TO FIND THE CENTER OF ANY CIRCLE

Suppose you have a circle without any center marked. (See Figure 23.) How are you going to find the exact center?

METHOD: Draw any triangle inside the circle, making sure that the approximate location of the center comes inside the triangle. This is the triangle ABC in the figure. Now erect perpendicular bisectors on A-B and A-C by the method given in Number 1, above. These two lines meet in a point which is the exact center of the circle. This is the point O.

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