« ΠροηγούμενηΣυνέχεια »
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.
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.
Secondly, let E be without the circumference of the circle.
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.
(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
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
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.
BA : AC= BD : DC.
[1. 8. and the alternate angle DAC= ACE,
[1. 8. but
AEC = ACE,
And since AD is parallel to CE,
[ıy. 2. but by hypothesis
BA: AC=BD : DC. Therefore
LE= AC, whence
BAD = CAD.
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.
In the triangle IB, ket FC be bisected by AD.