3D Geometry Proof by Contradiction /Contrapositive (high school) Could someone evaluate my work?
A plane perpendicular to one of 2 parallel lines is perpendicular to the other line also.
My two column proof so far:


*

*Let AB || CD and AB be perpendicular to plane p. Reason: Given

*Assume by way of contradiction that CD is not perpendicular to p. Reason: Assumption

*Let CX be drawn so that CX is perpendicular to p at point X. Reason: Assumption

*AB and CD form plane q. Reason: Two lines determine a plane (analogy to two points determine a line)

*CX and CD are both parallel to AB Reason: Steps 1,2 and 4

*CX = CD. Reason: ? 


I just know at the last step I have some kind of a contradiction which leads me to conclude that CD is perpendicular to p. Would I also have to use the Fifth postulate (parallel postulate) somewhere? 
 A: The "plane" and "line" directions determine each other uniquely when perpendicular. If plane $P$ and line $l$ are perpendicular, then any plane $P'$ parallel to $P$ is perpendicular to any line $l'$ parallel to $l$.  You are asked to prove the case of one plane $P$ and two parallel lines $l$ and $l'$.
Any proof that two things are perpendicular requires some known set of conditions under which perpendicularity can be inferred (such as a definition of perpendicularity, that you would then hope to show is satisfied).  One definition of perpendicularity between a line and a plane is that $\ell \perp P$ if $\ell$ intersects the plane $P$ and is perpendicular to every line in $P$ through the intersection point of $\ell$ and $P$. 
There is a canonical line segment joining the intersections of $P$ with the two given parallel lines.  A diagram that involves the intersection point with the first line can be translated (pushed without rotation) along that line segment to make a parallel diagram based at the intersection point on the second line.  So when checking the criterion for the first line to be perpendicular to $P$, move any perpendicular line involved in the argument at the first intersection point, along the special segment to the second line's intersection point with $P$, and the 90-degree angle formed at the first point will be cloned at the second.  Whatever is true at the first location will be transported to the second.  That's the logic and, in essence, a complete proof.
Note that this has absolutely nothing to do with proof by contradiction or contrapositive.  You have a quite affirmative way of transferring things that are true for one line (in relation to the plane of interest) to the other. 
