# Geometry- help me plz..

• Oct 30th 2006, 09:53 AM
jenjen
Geometry- help me plz..
1) Suppose we are given triangleABC and a point D in the interior of this triangle, and let E be any point in the same plane except D. What general conclusion about the intersection of triangleABC and [DE seems to be true?

[DE = ray

2) Let angleABC be given. Prove that there is a point D on the same plane such that D and A lie on opposite sides of BC, but d(A,B) = d(D,B) and
l angleABC l = l angleDBC l. (hint: this uses both the Ruler and Protractor Postulates).

Thankkk youu so much in advance.
• Oct 30th 2006, 11:09 AM
Plato
These answers may not exactly fit the definitions and axioms you are using. However, you should be able to adapt these to fit your text material.

The first is true. The interior of a triangle is bounded and rays are not. Hence the ray will intersect the triangle. In higher mathematics this is a version of the Jordan curve theorem. It can be very easy or very difficult to prove, it just depends upon the axioms you have. I have even seen it taken as an axiom itself.

As for the second, there are as many different so-called ‘protractor axioms’ as there textbooks that contain them. So what I can do is give you only an outline of a proof.
The $\displaystyle \overleftrightarrow{BC}$ has an A-side and a non A-side in the non A-side by the protractor axiom or a angle construction theorem there is a point H such the angle $\displaystyle \angle CBH$ is congruent to $\displaystyle \angle CBA$. By the ruler axiom on the line $\displaystyle \overleftrightarrow{BH}$ there is a coordinate system so that the coordinate of B is 0 and the coordinate of H is positive. Thus on the ray $\displaystyle \overrightarrow{BH}$ there is a point, D, such that $\displaystyle \left| {\overline {BA} } \right| = \left| {\overline {BD} } \right|.$
• Oct 31st 2006, 04:39 PM
jenjen
Thanks!