N. B. A segment of a line is a part cut off from a line; the segments spoken of above are to be reckoned from the circumference to the point of intersection and again from the point of intersection to the circumference.

Let AB, CD intersect in E.

Then shall AE. EB = CE. ED.

First, let E lie within the circumference of the circle.

Secondly, let E be without the circumference of the circle.

Join AC, BD.

Then EBD-ACE, since each is supplemen- [11. 15, Cor. 2. tary to ABD, and E is common to the triangles AEC, DEB.

Therefore these triangles are equiangular and similar.
Therefore AE: EC=DE : EB,

whence

AE, EB = CE, ED.

COR. Let E

[IV. 3.

be turned about E so as to become more remote from EC. The points A and B will continually approach one another and at last coincide. We shall then Hare

AE= 33 and 45. E3 = LE2.

* 4% de the agent at £

In the triangles CEA, AED, the angle at E is common; and the angle DAE= ACE, since EA touches the circle.

[II. 16.

If one angle of a triangle be bisected by a straight line which cuts the base, the sides of the triangle shall be to one another as the segments of the base.

In the triangle ABC, let A be bisected by AD.

Produce BA, and through C draw CE parallel to DA

meeting BA produced in E.

Then the exterior angle BAD = AEC,

[1. 8.

And since AD is parallel to CE,

BA: AE=BD: DC.

Therefore

BA: ACBD : DC.

[Iv. 2.

Conversely, if BA: AC = BD : DC, DA bisects BAC.

The same construction being made

If the exterior angle of a triangle be bisected by a straight line which cuts the base, the sides of the triangle shall be to one another as the segments of the base.

In the triangle ABC, let F4C be bisected by AD.

Then shall

BA:AC=BD: DC.

Through draw CE parallel to DA.

Then as in the preceding proposition AE=AC.

And

BA:AE=5D: DC

[IV. 2.