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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 Elie within the circumference of the circle.
Join AC, BD
Then

CAB = CDB, being on the same arc CB,

[11. 14. and AEC= DEB.

[I. 2. Therefore the triangles AEC, DEB are equiangular and similar. Therefore AE: EC=DE : EB,

(IV. 3. whence

AE. EB= CE. ED.

E

BD

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 B: E('= DE: EB,

(IV. 3. hence

E, ER = CE. ED. Chr. Lett be turned about E so as to become more ratite the At The points and B will continually Appalach otie sucher and at last cineide. We shall then tar

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

[II. 16. Therefore the triangles CEA, AED are equiangular and similar. Whence CE : EA = AE:ED,

[IV. 3. and therefore CEED = AE

THEOREM XIX.

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.

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

BA : AC= BD : DC.
Produce BA, and through C draw CE parallel to DA
meeting BA produced in E.
Then the exterior angle BAD = AEC,

[1. 8. and the alternate angle DAC= ACE,

[1. 8. but

BAD= DAC.

Therefore

AEC = ACE,

whence

AC= AE.

And since AD is parallel to CE,
BA: E= BD : DC.

[IV. 2.
Therefore BA: AC=BD: DC.
Conversely, if BA: AC=BD:DC, DA bisects BAC.
The same construction being made
BA : AE= BD: DC,

[ıy. 2. but by hypothesis

BA: AC=BD : DC. Therefore

LE= AC, whence

BAD = CAD.

THEOREN XX.

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

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In the triangle IB, ket FC be bisected by AD.
Then shall BI:d=BD:DC.
Through drar (E parallel to DA.
Then as in the preceding proposition !E= AC.

BI: E=BD:D

(IV. 2.

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