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

Proof

 * Euclid-XI-19.png

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 :
 * $DE$ 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.

Hence the result.