Proof: By Euclid

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.

fig19e

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.
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