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Show that the proposition P :

There exists a pair of straight lines that are at constant distance from each other.

is equivalent to the Parallel Postulate Q :

If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough.

I tried to prove $Q \rightarrow P$ then $\neg Q \rightarrow \neg P$. But for the second part, I can do nothing because as soon as the postulate is supposed to be untrue, the equivalent relation between angles no more exist, therefore it's hard to get congruent triangles as I used to do.

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1 Answer 1

up vote 1 down vote accepted

Just a hint:

You'd rather try $P\to Q$ instead, showing that the equidistant line is the only line on a given point which doesn't intersect the original line. Then, prove $\lnot Q$ implies that there are more lines that don't intersect.

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