Unizor - Geometry3D - Lines and Planes - 2 Lines Perpendicular to a Plane

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Theorem 1
If one of two parallel lines is perpendicular to some plane, the other one is perpendicular as well.

Proof
Assume that a and b are parallel lines and plane γ is perpendicular to a:
a ∥ b;
a⊥γ.
We will prove that b⊥γ.
Let the base of line a on plane γ (that is, their intersection a∩γ) be point A and the intersection b∩γ be point B. The latter intersection must exist because, otherwise, line b would be parallel to plane γ, and line a, as parallel to b, would also be parallel to γ, which contradicts their perpendicularity.
Lines a and b lie in the same plane because they are parallel, let's call this plane δ. In plane δ lines a and b are parallel and line AB (lying in both planes γ and δ) is perpendicular to a (since a⊥γ). Therefore, according to theorems of plane geometry, AB⊥b.
Since line AB belongs to plane γ and passes through intersection point B of b and γ, we have one line on γ that b is perpendicular to.
To complete the proof, we need a second line on γ that is perpendicular to b.
Let's draw two parallel lines c and d in plane γ through, correspondingly, bases A and B. We know that a⊥c (since a⊥γ). Compare an angle formed by a pair of lines a and c with an angle formed by a pair of lines b and d. The former is the right angle and both angles have correspondingly parallel sides. Therefore, these angles are congruent and the latter is the right angle as well.
So, the line d is the second line on plane γ that b is perpendicular to. Together with already proven AB⊥b, this is a sufficient condition for line b to be perpendicular to plane γ.
End of proof.

Theorem 2
If two lines are perpendicular to the same plane, they are parallel to each other.

Proof
Assume that a and b are two lines and plane γ is a plane perpendicular to both of them:
a⊥γ;
b⊥γ.
We have to prove that a ∥ b.
Let points A and B are, correspondingly, bases of perpendiculars a and b:
A=a∩γ,
B=b∩γ.
Construct a line b' that passes through point B and is parallel to line a. According to Theorem 1 above, b'⊥γ. If b and b' are not one and the same line, we have two perpendiculars to plane γ at the same point B, which we have proven before is impossible, the perpendicular to a plane at a point on it is unique.
Therefore, b=b' and, therefore, a∥b.
End of proof.

Theorem 3
If from a point M lying outside plane γ two different lines, a and b, are constructed in such a way that a⊥γ and b∩γ≠∅, then line b is NOT parallel to line a and is NOT perpendicular to plane γ.

Proof
Obviously, a and b are NOT parallel because they intersect at point M.
If b⊥γ, lines a and b would have been parallel (see Theorem 2 above), which we have just rejected.
So, b is NOT perpendicular to γ. Actually, it is slant or oblique line relative to plane γ.
End of proof.