(related to Proposition: Prop. 11.19: Common Section of Planes Perpendicular to other Plane is Perpendicular to that Plane)

- For let the two planes $AB$ and $BC$ be at right angles to a reference plane, and let their common section be $BD$.
- I say that $BD$ is at "right angles to the reference plane":bookofproofs$2212.

- For (if) not, let $DE$ also have been drawn from point $D$, in the plane $AB$, at right angles to the straight line $AD$, and $DF$, in the plane $BC$, at right angles to $CD$.
- And since the plane $AB$ is at right angles to the reference (plane), and $DE$ has been drawn at right angles to their common section $AD$, in the plane $AB$, $DE$ is thus at "right angles to the reference plane:bookofproofs$2212 " [Def. 11.4] .
- So, similarly, we can show that $DF$ is also at "right angles to the reference plane":bookofproofs$2212.
- Thus, two (different) straight lines are set up, at the same point $D$, at "right angles to the reference plane":bookofproofs$2212, on the same side.
- The very thing is impossible [Prop. 11.13].
- Thus, no (other straight line) except the common section $DB$ of the planes $AB$ and $BC$ can be set up at point $D$, at "right angles to the reference plane":bookofproofs$2212.
- Thus, if two planes cutting one another are at right angles to some plane then their common section will also be at right angles to the same plane.
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"