Existence of Right Angle in Hilbert Axioms 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?

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:


*

*$c=a$. Then angle $AP$ is supplementary to $AQ$ and they are congruent. Hence they are right angles.

*$c\in A$. 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.

*$c\in A^*$. 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.

