Question 991959
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Cannot be done with the information given.  Consider the following diagram that fits the given information.


*[illustration Counter_Example_For_Geometry_Proof.jpg]


The point is that it is possible to construct an angle DBC with a ray BE that does not bisect the angle.  Hence it is impossible to prove that it does, in general, bisect the angle.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \