3D geometry proof help (high school) My textbook is very different from regular high school textbooks because I go to a Christian academy. No one can help me though I was told that there are real math experts here. I need someone to proofread my work. If there is an alternative way please show me, it doesn't have to follow this strange format that my book requires. 
Statement to prove:
Let m be a plane parallel to plane n. Suppose that (line) AB is perpendicular to m, then (line) AB is also perpendicular to n. 
I wrote:


*

*Let BT || AP where BT is a line on plane n and AP is a line on plane m. 
Reason: Given

*AB is perpendicular to m
Reason: Given

*Suppose by way of contradiction that BT and AB intersect at a point P
Reason: Lines intersect at exactly one point

*But, plane m is parallel to BT 
Reason: Law of Contradiction

*AB is perpendicular to n
Reason: Law of Deduction 


This is not alright. First $A$ may not belong to $m$, nor $B$ to $n$. Second, even if planes $m$ and $n$ are not parallel, $BT$ and $AP$ may not intersect. Still, the general reasoning is good and the solution could be fixed easily:


*

*Set points $A' = AB \cap n$ and $B' = AB \cap m$.

*Assume that $m \not\parallel n$. This implies that they intersect at some line (and of course $m \neq n$).

*Let $P$ be any point such that $P \in m \cap n$.

*All segments $A'B'$, $A'P$ and $B'P$ lie in a common plane (the one that contains $A'$, $B'$ and $P$), however, $A'B' \perp A'P$ and $A'B' \perp B'P$.


I will leave the rest of the details to you. Good luck!
