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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:

  1. Let BT || AP where BT is a line on plane n and AP is a line on plane m. Reason: Given
  2. AB is perpendicular to m Reason: Given
  3. Suppose by way of contradiction that BT and AB intersect at a point P Reason: Lines intersect at exactly one point
  4. But, plane m is parallel to BT Reason: Law of Contradiction
  5. AB is perpendicular to n Reason: Law of Deduction
Person
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2 Answers2

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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!

dtldarek
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  • I see your reasoning, it makes sense. The last part looks like you're saying that the line A'B' is perpendicular to itself. Thanks for the tips – Person Mar 14 '13 at 08:17
  • @Person I am saying that $A' = B'$, beware, there are still missing pieces there. – dtldarek Mar 14 '13 at 08:37
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Did you know that alternate interior angles of parallel lines are congruent, or that the angle formed by parallel lines cutting a transversal is congruent? I don't remember the exact name, but if you know this, you don't need an indirect proof.

cuabanana
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