Existence of Right Angle in Hilbert Axioms

Question

Hilbert, in Foundations of Geometry briefly mentions that the existence of right angles is a corollary to the supplementary angle theorem. (i.e. If two angles are congruent, then their supplementary angles are congruent).

How does existence of right angles follow from this?

Answer 1

Well, it's not an immediate conclusion from this fact. However, this fact is used in the proof. The proof goes as follows:

Take a line $L$ and a point $p$ not lying on $L$. Next take a point $a\in L$ and choose a ray $A$ with origin $a$ which is contained in $L$. Let $P:=\overrightarrow{ap}$ and let $M$ be a halfplane with boundary $L$, to which point $p$ belongs. Next lay off a ray $Q$ with origin $a$ contained in the halplane $M^*$ (i.e. halfplane complementary to $M$) such that $AQ\equiv AP$. Next take a point $q\in Q$ such that $aq\equiv ap$. Since $p\in M$ and $q\in M^*$, the segment $\overline{pq}$ and line $L$ have a point in common. Call it $c$. Now we have three cases:

1. $c=a$. Then angle $AP$ is supplementary to $AQ$ and they are congruent. Hence they are right angles.
2. $c\in A$. Then by SAS $\triangle cap\equiv\triangle caq$. From this congruence we get $\angle acp\equiv \angle acq$ and these angles are supplementary. Hence they are right angles.
3. $c\in A^*$. Here we use the aforementioned fact. $AP,A^*P$ and $AQ,A^*Q$ are pairs of supplementary angles and $AP\equiv AQ$. So $\angle caq=A^*Q\equiv A^*P=\angle cap$. Again by SAS $\triangle cap\equiv\triangle caq$ and the rest is the same as in case 2.