1. From point B as centre, with radius BC, describe arc CD.

2. From point A as centre, with radius AC, describe arc CE.

A.

E

3. Draw the line CE, and it will be perpendicular to the given straight line AB.

Another Method.

1. Take any point D in the given line AB towards A. Join DC, and bisect it in E (Pr. 1).

2. From E as centre, with ED as radius, describe an arc cutting AB in F.

3. Draw the line CF, and it will be perpendicular to the given straight line AB.

Problem 4.

To bisect a given angle BAC.

1. From point A as centre, with any radius, describe an arc DE

B

VF

A

2. From point D as centre, with any radius, describe the arc FG; and from point E as centre, with the same radius, intersect the arc FG in the point H.

3. Draw the straight line AH, and it will bisect the given angle BAC.

Problem 5.

To trisect a given right angle ABC, that is, to divide it into three equal angles.

1. From point B as centre, with any radius, describe an arc DE.

2. From centres D and E, with the same radius, cut arc DE in G and F.

3. Draw the straight lines BF and BG, and the given right angle ABC will be trisected, that is, divided into three equal angles.

NOTE. It is only the right angle which (strictly speaking) can be trisected by a plane geometrical construction.

Problem 6.

To construct an angle of 45° at point B on a given line AB.
1. At point B, erect a perpendicular to AB (Pr. 2).
will then be a right angle, that is, an angle of 90°.

ABC

2. Bisect the right angle ABC (Pr. 4) by the line BD. Then DBA is the required angle.

NOTE.

In the same manner, an angle of 221° may be constructed by bisecting the angle DBA.

Problem 7.

To construct an angle of 60°, 30°, or 15° at point B on a given line AB.

1. With centre B, and any radius, describe an arc cutting AB in C.

2. From centre C, with the same radius, cut the arc in D.

3. Draw the line DB. Then DBA is the required angle of

60.

4. Bisect the angle DBA by the line EB (Pr. 4.) Then EBA is the required angle of 30°.

5. Bisect the angle EBA by the line FB (Pr. 4). Then FBA is the required angle of 15°.

NOTE 1.-In the same manner, an angle of 71° may be constructed by bisecting the angle FBA.

NOTE 2.-The radius of a circle is one-sixth of its circumfer-
It is on this principle that the angle DBA is 60°.

ence.

NOTE 3.-By means of this problem, we might construct other angles. Thus, if DBA be trisected by drawing lines to B, we obtain an angle of 20°. Bisect that, and we obtain an angle of 10°. Again, if angle FBC be trisected, by drawing lines to B, we might obtain an angle of 5°.

Problem 8.

To draw a line parallel to a given line AB, at a given distance from it, as CD.

1. Take any two points, E and F in the line AB, and with E and F as centres, and radius CD, describe arcs above the line.

2. Draw the line GH tangential to, or touching the arcs. Then the line GH will be parallel to the given line AB, and at the given distance CD from it.

Problem 9.

To draw a line parallel to a given line AB, through a given point C.

1. Take any point D in the given line AB towards A as centre; and from. D, with radius DC, describe an arc cutting AB in E.

2. From C as centre, with the same radius, describe an arc DF, and make arc DF equal to arc CE.

3. Draw the line FC, and it will be parallel to the given line AB.

Problem 10.

From a given point D, in the line AB, to make an angle equal to the given angle C.

1. From point C as centre, with any convenient radius, describe an arc EF.

2. From point D, with the same radius, describe the arc GH.

3. From point G, and the distance EF, cut off GK equal to EF.

4. Draw the line DK, and the angle KDB will be equal to the given angle C.