1. ## Proof

I have to prove the following:

Prove euclidean parallel postulate iff parallel lines are equidistant from one another. My hint is this:

Use the converse to Theorem 4.1 since it can be proved assuming Neutral geometery + euclidean parallel postulate. " Parallel lines are equidistan from one another" if the length of all perpendicular segments from a point on line "dropped" to the other line are congurent[equal in length].

2. Originally Posted by Nichelle14
Prove euclidean parallel postulate iff parallel lines are equidistant from one another
Please define for me in rigorous geometric terms what it means that two lines are equidistant.

3. the best I can do for that is say that they are the same measurement apart. I don't think I can give you an answer. Maybe someone else can.

4. Observe, the picture.
As I understand, the equidistant postulate states that,
AB is perpendicular to line AD
DC is perpendicular to line BC
Also, equidistance-meaning AB=DC
Note, triangle ABD and triangle BDC
They satisfy Hypotenuse-Leg. Thus, they are congruent. Thus, <ADB=<BDC
Thus the parallel postulate is proved.

Note, the congrunce of hypotenuse-leg is independent of parallel postulate, If it were not then you cannot use it.