Construction of Straight Line Perpendicular to Plane from point not on Plane

Theorem

In the words of Euclid:

From a given elevated point to draw a straight line perpendicular to a given plane.

(The Elements: Book $\text{XI}$: Proposition $11$)

Proof

Let $A$ be the given elevated point.

Let the given plane be identified as the plane of reference.

It is required that a straight line be drawn perpendicular to the plane of reference.

Let any arbitrary straight line $BC$ be drawn in the plane of reference.

From Proposition $11$ of Book $\text{I} $: Construction of Perpendicular Line:

Let $AD$ be drawn from $A$ perpendicular to $BC$.

If $AD$ is perpendicular to the plane of reference, then the construction is complete.

Otherwise, let $DE$ be drawn from $D$ perpendicular to $BC$ and in the plane of reference.

From Proposition $12$ of Book $\text{I} $: Perpendicular through Given Point:

Let $AF$ be drawn from $A$ perpendicular to $DE$.

From Proposition $31$ of Book $\text{I} $: Construction of Parallel Line:

Let $GH$ be drawn from the point $F$ parallel to $BC$.

We have that $BC$ is perpendicular to each of $DA$ and $DE$.

Therefore from Proposition $4$ of Book $\text{XI} $: Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane:

$BC$ is perpendicular to the plane through $ED$ and $DA$.

We have that $GH$ is parallel to $BC$.

But from Proposition $8$ of Book $\text{XI} $: Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane:

$GH$ is also perpendicular to the plane through $ED$ and $DA$.

Therefore by Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:

$GH$ is perpendicular to all the straight lines which meet it and are in the plane through $ED$ and $DA$.

But $AF$ meets $GH$ and is in the plane through $ED$ and $DA$.

Therefore $GH$ is perpendicular to $FA$.

So $FA$ is perpendicular to $GH$.

But $FA$ is also perpendicular to $DE$.

Therefore $FA$ is perpendicular to both $GH$ and $DE$.

Therefore from Proposition $4$ of Book $\text{XI} $: Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane:

$AF$ is perpendicular to the plane through $DE$ and $GH$.

But the plane through $DE$ and $GH$ is the plane of reference.

Therefore $AF$ is perpendicular to the plane of reference.

Thus $AF$ is the required line.

$\blacksquare$

Historical Note

This proof is Proposition $11$ of Book $\text{XI}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions