Equivalents to the Euclidean Postulate

In this section we work within neutral to prove that a number of different statements are equivalent to the Euclidean (EPP). This has historical importance. We noted before that ’s Fifth Postulate was quite different in tone and complexity from his first four postulates, and that Euclid himself put off using the fifth postulate for as long as possible in his development of geometry. The is rife with efforts to either prove the fifth postulate from the first four, or to find a simpler postulate from which the fifth postulate could be proved. These efforts gave rise to a number of candidates that, in the end, proved to be no simpler than the fifth postulate and in fact turned out to be logically equivalent, at least in the context of neutral geometry. Euclid’s fifth postulate indeed captured something fundamental about geometry.

We begin with the converse of the Alternate Interior , as follows:

If two parallel lines n and m are cut by a l , then alternate interior angles are congruent.

We noted earlier that this statement was equivalent to the Euclidean Parallel Postulate. We now prove that.

Theorem: The converse of the Alternate Interior Angles Theorem is equivalent to the Euclidean Parallel Postulate.

~ We first assume the converse of the Alternate Interior Angles Theorem and prove that through a given l and a P not on the line, there is exactly one line through P parallel to l. We have already proved that one such line must exist: Drop a from P to l; call the foot of that perpendicular Q. Let . Now, through point P use the protractor postulate to construct a line m perpendicular to t. The Alternate Interior Angles Theorem guarantees that since the alternate interior angles are all right angles, l and m are parallel. So it all comes down to proving that there is only one such line. To this end, suppose there were another line mN through P parallel to l. By the converse to the Alternate Interior Angles Theorem, alternate interior angles formed by l, mN, and t must be equal, and since t z l we must have mN z t as well. But then m and mN would both be to t though point P, which would violate the uniqueness of rays forming a right with t through a given point (Protractor Postulate). Thus, m = mN.

To prove the converse, assume that the EPP holds and that l and m are parallel lines cut by a transversal t at points Q and P, respectively. We prove that alternate interior angles are congruent. Suppose for contradiction that alternate interior angles p1 (at point P) and p2 (at point Q) are not congruent, and WLOG that :(p1) > :(p2). Using the protractor postulate, create ray on the other side of t from p2 such that :(pRPQ) = :(p2). Then from the above theorem, since pRPQ and p2 are alternate interior angles, it follows that lines l and are parallel. But that means there are two lines through P parallel to line l, contradicting the parallel postulate. So p1 and p2 are congruent. € Theorem: Euclid’s Postulate V is equivalent to the Euclidean Parallel Postulate.

~ First we assume EPP and prove from it Postulate V. Suppose l and m are two lines cut by a transversal t at points P and Q respectively in such a way that the interior angles on one side of t have measures adding to less than 180. Using the Protractor Postulate, construct through point P a line n such that the interior angles formed by n and m (on the same side of t as before) add up to 180.

Now a Corollary to the Alternate Interior Angles Theorem, states that if two lines are cut by a transversal such that the interior angles on one side of the transversal are supplementary, the lines are parallel. Thus n and m are parallel, and by EPP n is the only line through P that is parallel to m. Thus l is not parallel to m, and must meet. They must meet on the side of t we have been working on or the they form would have angle measure greater than 180.

To prove the converse, let l’ be a line and P a point not on the line. We have demonstrated how we can create a line through P parallel to l’ by using perpendiculars, so we will let l be such a line, created by dropping a perpendicular from P to l’ with foot Q, then forming line l perpendicular to at point P. We now prove there can be no other lines through P parallel to l’. Suppose m is such a line. We know that m cannot be perpendicular to at P since l is, and so the angle it makes with t must have measure less than 90 on one side of t. On that side, the sum of the interior angles formed by m and l’ add to less than 180, and so l’ and m must meet on that side. Thus l’ and m cannot be parallel. € Theorem: The following statements are each equivalent to the Euclidean Parallel Postulate (EPP):

1. If l and l’ are parallel lines and is a line such that t intersects l, then t also intersects l’.

2. If l and l’ are parallel lines and t is a transversal such that , then .

3. If l, m, n, and k are lines such that , then either m = n or .

4. If l is parallel to m and m is parallel to n then either l = n or .

~ To prove statement 1 from EPP, let l and l’ be parallel lines and be a line such that t intersects l at point P. By EPP, there can only be one line through point P that is parallel to l’, so t must intersect l’.

To prove the converse, let l’ be a line and P a point not on l’. We have demonstrated how we can create a line through P parallel to l’ by using perpendiculars, so we will let l be such a line. We now prove there can be no other lines through P parallel to l’. Let be a line through point P. By statement 1, since t intersects l, it must also intersect l’, and so cannot be parallel to l’. Thus there can only be one line through P which is parallel to l’. To prove statement 2 from EPP, let l and l’ be parallel lines and t a transversal such that . The EPP is equivalent to the converse of the Alternate Interior Angle Theorem, so since l and l’ are parallel, alternate interior angles p1 and p2 must be congruent, and since p1 is a , p2 must be a right angle, so that .

To prove the converse, let l’ be a line and P a point not on l’. We have demonstrated how we can create a line through P parallel to l’ by using perpendiculars, so we will let l be such a line, created by dropping a perpendicular from P to l’ with foot Q, then forming line l perpendicular to at point P. We now prove there can be no other lines through P parallel to l’. Suppose m is a line through P parallel to l’. Now is a transversal for m and l’ that is perpendicular to l’ at Q, and so by statement 2 it must also be perpendicular to m at P. But this makes l and m both perpendiculars to at P, contradicting the Protractor Postulate. Thus there cannot be a second parallel m. To prove statement 3 from EPP, we note that we have already proved that statements 1 and 2 above are equivalent to EPP, so we can use them. So let l, m, n, and k be lines such that , and suppose that . In particular, since l and k are parallel and m intersects k, it must also intersect l by statement 1. Moreover, by statement 2, since as well. But then, since and , if we view l as a transversal for m and n we have alternate interior angles congruent, and so .

To prove the converse, let l’ be a line and P a point not on l’. We have demonstrated how we can create a line through P parallel to l’ by using perpendiculars, so we will let l be such a line, created by dropping a perpendicular from P to l’ with foot Q, then forming line l perpendicular to at point P. We now prove there can be no other lines through P parallel to l’. Suppose m is a line through P parallel to l’. Drop a perpendicular from Q to m with foot F. Now and , so by statement 3 either , or . Since and clearly share point Q, we must have . Now since can only intersect m once, F = P. But this contradicts the Protractor Postulate, since l and m are both perpendicular to at P.

To prove statement 4 from EPP, suppose that l is parallel to m and m is parallel to n , and that l and n are not equal. We prove . Suppose for contradiction that l and n intersect. Then since l is parallel to m, by statement 1, n and m intersect, a contradiction. Thus l and n cannot intersect and are therefore parallel.

To prove the converse, let l’ be a line and P a point not on l’. We have demonstrated how we can create a line through P parallel to l’ by using perpendiculars, so we will let l be such a line, and suppose m is a second line through P parallel to l’. But then l is parallel to l’ and l’ is parallel to m so by statement 4, l is parallel to m, which cannot be since they share the point P. Thus no such line m can exist. €

Whew!

Our next Tour de Force is to prove that the EPP is equivalent to the claim that the angle sum for every triangle is 180. One direction of the equivalence is easy; the other is messier and requires some properties of real numbers. Theorem: The statement that the angle sum for any triangle is 180 is equivalent to the EPP.

~ We’ll do the easy direction first; let be any triangle, and through point C construct a line parallel to .

Now, (I’ll leave you to fill in the betweenness details necessary to make all that work) and by the converse to the Alternate Interior Angles Theorem (equivalent to EPP), and . Substitution gives .

OK, now for the harder direction. We prove it by contradiction, so we will actually show that if there is a line and external point through which many parallels exist, then there is a triangle whose angle sum is less than 180.

We need a lemma first, one vaguely reminiscent one part of the proof of the Saccheri-Legendre Theorem. Lemma: Let be a segment and let points QN be a point such that pPQQN is a right angle. Then for every g > 0 there exists a point T on such that .

~ Choose a point PNon the same side of as QNsuch that pQPPN is

a right angle (i.e. ). On ray find the point T1 with

QT12 = PQ. Next, find the point T with T1T2 = PT1. Continue in this way, choosing Tn so that Q*Tn-1*Tn and PTn-1 = TnTn-1. This creates a series of isosceles with base angles so that

. Moreover, each of the Tn’s are interior to angle pQPPN. This gives us: , But . Thus the sum of the measures of the angles we are forming with V must be less than 90. Now suppose for contradiction that for every . By the Archimedean Property of Real Numbers there is an n for which , so that , which contradicts our previous conclusion about this sum. Thus there must be some for which and so some for which . € Now the other direction of the main theorem:

Let l be a line and P a point not on l through which there is more than one line parallel to l. Build a line m in the usual way by dropping a perpendicular with foot Q from P to l and creating a line m perpendicular to through P. Now pick another line m’ through P parallel to l. The angle m’ makes with is not of measure 90 so on one side or the other of makes an angle less than 90 with m’. On that side pick a point S on m’ so that S is on the same side of m as Q. Choose a point R on m such that R is on the same side of as S. Finally, using our lemma above, choose a point T on l such that T is on the same side of as S, and . We will show that the angle sum for is less than 180.

First note that T was constructed to be on the S side of and that if it were not on the Q side of then would have to intersect m’ = by Separation. This cannot happen since m’ is parallel to l. So T is interior to pQPS, and we have that by additivity of angles. Moreover, we chose S so that S is interior to pQPR, so we have . This gives us: or or €