(Edited by Dionysius Lardner, 1855)

Proposition IX. Problem.

[Euclid, ed. Lardner, 1855, on `Google Books`]

(70) | To bisect a given rectilinear angle (B A C). |

Take any point D in the side A B, and from A C cut off A E equal to A D (III), draw D E, and upon it describe an equilateral triangle D F E (I) at the side remote from A. The right line joining the points A and F bisects the given angle B A C.

Because the sides A D and A E are equal (const.), and the side A F is common to the triangles F A D and F A E, and the base F D is also equal to F E (const.); the angles D A F and E A F are equal (VIII), and therefore the right line bisects the given angle.

Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)

Next: Proposition 10

Previous: Proposition 8

This proposition in other editions:

By this proposition an angle may be divided into 4. 8, 16 &c. equal parts, or, in general, into any number of equal parts which is expressed by a power of

two.It is necessary that the equilateral triangle be constructed on a different side of the joining line D E from that on which the given angle is placed, lest the vertex F of the equilateral triangle should happen to coincide with the vertex A of the given angle; in which case there would be no joining line F A, and therefore no solution.

In these cases, however, in which the vertex of the equilateral triangle does not coincide with that of the given angle, the problem can be solved by constructing the equilateral triangle on the same side of the joining line D E with the given angle. Separate demonstrations are necessary for the two positions which the vertices may assume.

1. Let the vertex of the equilateral triangle fall within that of the given angle.

The demonstration already given will apply to this without any modification.

2. Let the vertex of the given angle fall within the equilateral triangle.

The line F A produced will in this case bisect the angle; for the three sides of the triangle D F A are respectively equal to those of the triangle E F A. Hence the angles D F A and E F A are equal (VIII). Also, in the triangles D F G and E F G the sides D F and E F are equal, the side G F is common, and the angles D F G and E F G are equal; hence (IV) the bases D G and E G are equal, and also the angles D G A and E G A. Again, in the triangles D G A and E G A the sides D G and E G are equal, A G is common, and the angles at G are equal; hence (IV) the angles D A G and E A G are equal, and therefore the angle B A C is bisected by A G.

It is evident, that an isosceles triangle constructed on the joining line D E would equally answer the purpose of the solution.