|
|
|
|
|
|
|
|
|
Property of Angle Bisectors: What is this about?
|
| Buy this applet What if applet does not run? |
Angle bisectors divide the opposite side in the ratio of the adjacent sides. More accurately, if, in 
ABC, AD is an angle bisector of angle A, then
| AB/AC = DB/DC |
Note that the same holds also for the external angle bisectors.
| Buy this applet What if applet does not run? |
Proof
Assume the straight line through C parallel to AD meets AB in E. Then, first of all, AEC is isosceles:
 ACE = CAD = BAD = AEC.
|
Therefore, AE = AC, and the required proportion follows from the similarity of triangles BEC and BAD. (There is a less standard proof.)
This property of angle bisectors is one way to show that the three angle bisectors in a triangle meet in a point. The result is an immediate consequence of Ceva's theorem.
Last note: the converse theorem holds as a matter of course, because there is only one point on a given segment that divides it in a given ratio. Thus if a point divides the base of a triangle in the ratio equal to the ratio of the sides, it is bound to be the foot of the angle bisector from the apex.
| 29713871 |  |
|