Unizor - Geometry3D - Lines and Planes - Parallel Planes

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Prior to listening this lecture, we recommend to recall the main axioms and already proven theorems from previous lectures.

Definition
Two planes in three-dimensional space are called parallel if they do not intersect (that is, have no common points).

There are a few theorems related to parallel planes that we will prove in this lecture.

Theorem 1
(sufficient condition for two planes to be parallel)
Assume that we have two planes γ and δ.
Plane γ contains two intersecting straight lines aγ and bγ.
Plane δ contains two intersecting straight lines aδ and bδ.
These lines are correspondingly parallel, that is
aγ ∥ aδ and bγ ∥ bδ.
Prove that planes are parallel, that is γ ∥ δ.

Proof
Since aγ ∥ aδ, line aγ and plane δ are parallel. Similarly, since bγ ∥ bδ, line bγ and plane δ are parallel.
So, both intersecting lines aγ and bγ are parallel to plane δ.
Assume that planes γ and δ intersect. Their intersection must be some straight line c, according to Axiom 2 introduced earlier in the course. This line c belongs to both planes γ and δ.
Now we have a situation when line aγ is parallel to plane δ and lies in plane γ that intersects plane δ along a line c. According to Theorem 2 of the previous lecture about lines parallel to planes, the line of intersection c is parallel to line aγ.
Similarly, we have a situation when line bγ is parallel to plane δ and lies in plane γ that intersects plane δ along a line c. Therefore, c is parallel to line bγ.
As a result, line c in plane γ must be parallel to two lines, aγ and bγ, intersecting each other. This is not possible in the plane Euclidian geometry. Therefore, our assumption that planes γ and δ intersect is invalid, these planes must have no common points, that is they are parallel.
End of proof.

Theorem 2
Consider two parallel planes γ and δ (symbolically, γ ∥ δ). There is the third plane α that intersects both of them along lines aγ and aδ correspondingly:
γ∩α = aγ
δ∩α = aδ
Prove that aγ ∥ aδ.

Proof
The definition of parallel lines includes two requirements:
(a) lines must lie in the same plane; in our case both lines lie in plane α.
(b) they must not intersect; in our case they do not intersect since they belong to parallel planes.
End of proof.

Theorem 3
Consider two parallel planes γ and δ (symbolically, γ ∥ δ). There are two parallel lines a and b intersecting both these planes.
Let the points of intersection be as follows:
a∩γ = Aγ
a∩δ = Aδ
b∩γ = Bγ
b∩δ = Bδ
Prove that AγBγBδAδ is a parallelogram.

Proof
Line AγBγ and line AδBδ are parallel, according to Theorem 2 above.
Line AγAδ and line BγBδ are parallel, according a preposition of this theorem.
All four points lie in one plane since this is a condition of lines AγAδ and BγBδ being parallel.
Therefore, AγBγBδAδ is a quadrilateral with parallel opposite sides. According to one of the theorems of geometry of quadrilaterals on the plane, this is a parallelogram.


Theorem 4
(extension of Theorem 1 above)
Assume that we have two planes γ and δ.
Plane γ contains two rays originating from the same point Aγ:
AγBγ and AγCγ.
Plane δ contains two rays originating from the same point Aδ:
AδBδ and AδCδ.
These rays are correspondingly parallel and similarly directed, that is
AγBγ ∥ AδBδ and AγCγ ∥ AδCδ.
Prove that angles ∠BγAγCγ and ∠BδAδCδ lie in parallel planes (that is, γ ∥ δ) and are congruent.

Proof
The fact that γ ∥ δ is a subject of Theorem 1 in this lecture. So, it's already proven.
To prove congruence of angles, we will construct triangles where they belong to.
Choose points Bγ, Cγ, Bδ and Cδ such that AγBγ=AδBδ and AγCγ=AδCδ.
Now we have:
AγBγBδAδ is a parallelogram since AγBγ ∥ AδBδ and are equal in length (by construction).
Therefore,
AγAδ ∥ BγBδ and equal in length.
AγCγCδAδ is a parallelogram since AγCγ ∥ AδCδ and are equal in length (by construction).
Therefore,
AγAδ ∥ CγCδ and equal in length.
Based on the two statements above,
BγCγCδBδ is a parallelogram since BγBδ ∥ CγCδ and are equal in length.
Therefore, BγCγ ∥ BδCδ and equal in length.
Hence, two triangles, ΔAγBγCγ and ΔAδBδCδ are congruent by three sides, from which follows the congruence of angles ∠BγAγCγ and ∠BδAδCδ.
End of proof.