Common Section of Planes Perpendicular to other Plane is Perpendicular to that Plane

Theorem

In the words of Euclid:

If two planes which cut one another be at right angles to any plane, their common section will also be at right angles to the same plane.

(The Elements: Book $\text{XI}$: Proposition $19$)

Proof

Let $AB$ and $BC$ be planes which are perpendicular to the plane of reference.

Let $BD$ be the common section of $AB$ and $BC$.

It is to be demonstrated that $BD$ is perpendicular to the plane of reference.

Suppose $BD$ is not perpendicular to the plane of reference.

From the point $D$, let:

$DE$ be drawn in the plane $AB$ perpendicular to $AD$

$DF$ be drawn in the plane $AC$ perpendicular to $CD$

We have that:

$AB$ is perpendicular to the plane of reference

and:

$DE$ is in the plane $AB$ perpendicular to $AD$.

Therefore from Book $\text{XI}$ Definition $4$: Plane at Right Angles to Plane:

$DE$ is perpendicular to the plane of reference.

Similarly it can be proved that $DF$ is also perpendicular to the plane of reference.

Therefore from $D$ we have two straight lines have been set up perpendicular to the plane of reference on the same side.

This contradicts Proposition $13$ of Book $\text{XI} $: Straight Line Perpendicular to Plane from Point is Unique.

Hence the result.

$\blacksquare$

Historical Note

This proof is Proposition $19$ 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