How can the dihedral angle (i.e. the angle between two planes) be defined sans a coordinate system?

Question

Let $P$ and $Q$ two planes in the Euclidean $3$-dimensional space that intersect in a line $L$.

In the presence of a Cartesian coordinate system, the angle between $P$ and $Q$ can be defined as the angle between any two non-zero vectors $v$ and $w$ that are orthogonal to $P$ and to $Q$, respectively, and such that the angle between $v$ and $w$ is not obtuse. It can be shown with the tools of Analytic Geometry that this angle is the same regardless of the vectors $v$ and $w$ chosen from the many pairs of vectors that satisfy the requirements.

How can the angle between planes be defined in the absence of a Cartesian coordinate system, and without recourse to the tools of Analytic Geometry, but only to Hilbert's axioms? How can this definition be shown to be well-defined?

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I expect the definition to go along the following lines.

Choose any point $A$ on $L$, and let $S$ and $T$ denote the lines lying in $P$ and $Q$, respectively, that are perpendicular to $L$ and pass through $A$. Then the non-obtuse angle between $S$ and $T$ is defined to be the angle between $P$ and $Q$.

The real question that I'm interested in is: how can it be shown that this definition is well-defined and is independent of the choice of the point $A$?

Answer 1

Let $P$ and $Q$ be planes which intersect along the line $L$ and let $H,K$ be chosen halfplanes of $P$ and $Q$ respectively (with the boundary being the line $L$).

Let $o$ be a point in $L$ and let $A, B$ be rays of origin $o$ perpendicular to $L$ contained in $H, K$ respectively. We will show that the angle $AB$ does not depend on the choice of $o$.

So let $o'$ be a point distinct from $o$ and let $A',B'$ be rays of origin $o'$ perpedicular to $L$ contained in $H, K$ respectively. There exist points $a\in A, b\in B, a'\in A', b'\in B'$ such that $oa\equiv o'a', ob\equiv o'b'$. Since also $oa\parallel o'a', ob\parallel o'b'$ we get that $\square oaa'o'$ and $\square obb'o'$ are parallelograms (in fact rectangles). Hence $aa'\parallel oo'\parallel bb'$ and $aa'\equiv oo'\equiv bb'$ so (by transitivity of parallelism and congruence) $\square aa'b'b$ is a parallelogram and $ab\equiv a'b'$. By SSS congruence criterion $AB\equiv A'B'$.