Existence of Right Angle in Hilbert AxiomsHilbert, 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?

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Ezequiel Davidson · Accepted Answer

Step 1It&#039;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  ∈  L and choose a ray A with origin a which is contained in L. Let   P  :=            a      p        →   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   A  Q  ≡  A  P. Next take a point   q  ∈  Q such that   a  q  ≡  a  p. Since   p  ∈  M and   q  ∈      M    ∗  , the segment             p      q        ¯   and line L have a point in common. Call it c. Now we have three cases:Step 21.   c  =  a. Then angle AP is supplementary to AQ and they are congruent. Hence they are right angles.2.   c  ∈  A. Then by SAS   △  c  a  p  ≡  △  c  a  q. From this congruence we get   ∠  a  c  p  ≡  ∠  a  c  q and these angles are supplementary. Hence they are right angles.3.   c  ∈      A    ∗  . Here we use the aforementioned fact.   A  P  ,      A    ∗    P and   A  Q  ,      A    ∗    Q are pairs of supplementary angles and   A  P  ≡  A  Q. So   ∠  c  a  q  =      A    ∗    Q  ≡      A    ∗    P  =  ∠  c  a  p. Again by SAS   △  c  a  p  ≡  △  c  a  q and the rest is the same as in case 2.

Existence of Right Angle in Hilbert Axioms

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