Find the flaw in this proof of the parallel postulate.

[scamuicune]

Homework Statement

Given P not on line l, line PQ perpendicular to l at Q, line m perpendicular to line PQ at P, and point A ≠ P on m. Let ray PB be the last ray between ray PA and ray PQ that intersects l, B being the point of intersection. There exists a point C on l such that Q*B*C. It follows that ray PB is not the last ray between ray PA and ray PQ that intersects l, and hence all rays between ray PA and ray PQ meet l. Thus, m is the only parallel to l through P.

Homework Equations

Betweenness axioms
Definition of plane separation

The Attempt at a Solution

I'm pretty sure there's something not quite right with the statement that because Q*B*C, that QC is between rays PQ and PA. I feel like we're assuming something and that the flaw is in this statement.

 

[lippyka]

I think this statement should be something like, since Q*B*C, there is another point D on l such that QC is between rays PQ and PD. Then we can say that all rays between PA and PQ will intersect l. Then, to prove the last part of the statement, we can use the plane separation theorem which states that if two distinct lines have a common perpendicular, then they must be parallel. In this case, m is the common perpendicular since it is perpendicular to PQ. Thus, m is the only line parallel to l through P.