A

E

B

*47. The circumference of every circle is supposed to be divided into 360 equal parts, called degrees; each degree into 60 equal parts, called minutes; and so on.

48. The measure of any right-lined angle is an arc of a circle contained between the two lines which form that angle, the angular point being the centre.

Note. The angle is estimated by the number of degrees contained in the arc; whence a right angle is an angle of 90 degrees, or of the circumference.

PROBLEM I.†

To divide a given line AB into two equal parts.

This and the following definition are used only in Practical Geometry.

+ The demonstrations of most of these problems may be found in Euclid's Elements.

1. From the points A and B, as centres, with any distance greater than half AB, describe arcs cutting each other in n and m.

2. Through these points, draw the line nEm, and the point E, where it cuts AB, will be the middle of the line required.

PROBLEM II.

To divide a given angle ABC into two equal parts.

1. From the point B, with any radius, describe the arc AC.

2. And from AC, with the same, or any other radius, describe arcs cutting each other in n.

3. Through the point n draw the line Bn, and it will bisect the angle ABC, as was required.

PROBLEM III.

From a given point C, in a given right line AB, to erect a perpendicular.

CASE I. When the point is near the middle of the line.

1. On each side of the point C take any two equal distances Cn, Cm.

2. From n and m, with any radius greater than Cn or Cm, describe arcs cutting each other in s.

3. Through the point s, draw the line sC, and it will be the perpendicular required.

CASE II. When the point is at, or near, the end of the line.

1. Take any point o, and with the radius or distance oC, describe the arc mCn, cutting AB in m and C.

2. Through the centre o, and the point m, draw the line mon, cutting the arc mCn in n.

3. From the point n, draw the line nC, and it will be the perpendicular required.

1. From the point C, with any radius, describe the arc rnm, cutting the line AC in r.

2. With the same radius, and r as a centre, cross the arc in n; and from n in like manner, cross it in m.

3. From the points n and m, with the same, or any other radius, describe arcs cutting each other in s.

4. Through the point s, draw the line sC, and it will be the perpendicular required.*

PROBLEM IV.

From a given point C, out of a given line AB, to let fall a perpendicular.

CASE I. When the point is nearly opposite to the middle of the line.

1. From the point C, with any radius, describe the arc nm, cutting AB in n and m.

2. From the points n, m, with the same or any dius, describe two arcs cutting each other in s.

other ra

3. Through the points C, s, draw the line CGs, and CG will be the perpendicular required.

CASE II. When the point is nearly opposite to the end of the line.

* For another method of raising a perpendicular from any point in a given line, see Prob. XXXIX.

G

1. To any point m, in the line AB, draw the line Cm. 2. Bisect the line Cm, or divide it into two equal parts in the point n.

3. From n, with the radius nm, or nC, describe the arc CGm, cutting AB in G.

4. Through the point C, draw the line CG, and it will be the perpendicular required.

1. From A, or any other point in AB, with the radius AC, describe the arc CD.

2. And from any other point n, in AB, with the radius nC, describe another arc cutting the former in C, D.

3. Through the points C, D, drawn the line CGD, and CG will be the perpendicular required.

N. B. Perpendiculars may be more easily raised, and

C