Proculus\'s Axiom states that if I and l\' are parallel lines and t notequalto l

Question

Proculus's Axiom states that if I and l' are parallel lines and t notequalto l is a line such that t intersects l', then t also intersects ''. Prove the following theorem in Neutral Geometry. THEOREM: Euclidean Parallel Postulate is equivalent to Proculus's Axiom. PROOF: First assume EPP by. Let l, l', and t notequalto l be lines such that l||l' and t intersection theta by. We must prove that also intersects l' Suppose in the way of contradiction, that t intersection;' = theta. Then t||l' by definition of parallel lines. Let P elementof t intersection I. Now P is an external point for '', P lies on both I and, and both l and t are to l'. This contradicts EPP. Therefore the statement is proved. For the converse, we assume Proclus As Axiom by. Let l be a line and let P be an external point by. We must prove that there exists exactly line m such that lies on m and ||l. Drop a perpendicular from P to I and call the foot Q. Let m be the line such that P lies on m and m perpendicular PQ doubleheadarrow by Axiom. Then m||l by Theorem. Suppose there exists a line m' notequalto m such that P lies on m' ||l. By Proculus Axiom, it must be the case that m' intersection l notequalto theta. This contradicts the fact that m'||l, and thus EPP is proved.

Explanation / Answer

Euclidean Parallel Postulate implies Proclus's Axiom: Let l and l' be parallel lines. Let t l be a line such that t intersects l. Suppose that t does not intersect l'. So t || l'. Let P be the intersection of l and t. Then there are two distinct lines through P that are parallel to l', which contradicts the Euclidean Parallel Postulate. So t crosses l'. Assume Proclus's Axiom. Let l be a line and P an external point of l. Suppose that m is parallel to l and passes through P. Let n be a line that passes through P and is distinct from m. Then since n intersects m, n also intersects l by Proclus's Axiom. So there is only one parallel line through P that is parallel to l.

Assume the Euclidean Parallel Postulate. Let l and l' be lines, l || l', and t a transversal such that t l. Let P be the point at the intersection of l and t. Since t l, then t l. And since there is only one line through P that is parallel to l', it follows that t is not parallel to l', so t must intersect l'. Let Q be the point of intersection between l' and t. By definition of perpendicular lines, the angles at the intersection of l and t are right angles. By the Converse to the Alternate Interior Angles Theorem, the angles at the intersection of t and l' are also right angles. Thus, by definition of perpendicular lines, l' t. Assume Theorem 4.7.3 #2. Let l be a line and P a point external to l. Let t be the perpendicular dropped from P to l. Suppose that m and m' are two lines through P that are parallel to l. Then t m and t m'. Then the angles at the intersection of t and m are right angles, as are the angles at the intersection of t and m' (Def of Perp Lines). Then m = m' (Protractor Postulate). So there is only one line through P that is parallel to l.

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