Unizor - Geometry3D - Lines and Planes - Problems 3

Unizor - Creative Minds through Art of Mathematics - Math4Teens

We would like to address this construction problem in a very detailed way to exemplify the rigorousness required from mathematics, if we want to approach this subject on an advanced level.
This rigorousness requires to
(1) analyze the construction problem to properly plan the construction on a step by step level;
(2) present the construction steps and
(3) prove that this construction leads to required results.

Problem
Drop a perpendicular a to a given plane σ from a given point M outside of a given plane.

Analysis
We don't know yet how to drop a perpendicular to a plane from a point outside of it. But, introducing the concept of a line that is perpendicular to a plane (see "Line ⊥ Plane" lecture), we have constructed a plane perpendicular to a line at a given point on this line and also constructed a line perpendicular to plane at a given point on this plane.
Using the latter construction, we can choose any point N on a given plane σ (N∈σ), construct a line b perpendicular to a plane at this point (b⊥σ; N∈b) and, finally, construct line a parallel to line b, but passing through a given point M (a ∥ b; M∈a).
By the Theorem 1 from the lecture "2 Lines ⊥ Plane", this line a should satisfy the requirements of a problem - passing through point M and be perpendicular to plane σ.

Construction
Step 1. Choose any point N∈σ.
Step 2. Draw a perpendicular b to plane σ through point N on this plane as described in lecture "2 Lines ⊥ Plane".
Step 3. Draw a line a through point M parallel to line b as described in Exercise 1 of lecture "Line ∥ Line".

Proof of construction
Since line b is perpendicular to plane σ by construction and line a is parallel to line b, line a is perpendicular to plane σ.

Proof of uniqueness
Assume that we have two lines a and a' such that
a⊥σ, a∩σ=P
M∈a;
a'⊥σ, a'∩σ=P';
M∈a'.
' Draw a plane through intersecting lines a and a'. It will intersect plane σ on a line connecting P and P'.
Now lines MP and MP' are both perpendicular to line PP', which is impossible.
Therefore, points P and P' are one and the same, and the perpendicular from point M to plane σ is unique.