In Book I Proposition 10 of the Elements, Euclid performs the bisection (i.e. finding a midpoint) of a line segment. In the course of doing so, he uses Book I Proposition 4, the Side-Angle-Side Theorem, which proved with his controversial method of superposition. My question is, could Euclid have proven Book I Proposition 10 without relying on Book I Proposition 4, so that the result wouldn't depend on the method of superposition?

The reason I ask is that I think I've found a simpler proof of Euclid's Book I Proposition 2 (involving the transferring of distances), but it relies on bisection, so I'm hoping that I'm not implicitly using the method of superposition.

Any help would be greatly appreciated.

Thank You in Advance.

  • $\begingroup$ See Book I Proposition 9. Euclid could have made the reasoning for $n=2$ which would not resort to bissection. $\endgroup$ Dec 29, 2013 at 5:05
  • $\begingroup$ @DoktoroReichard If you trace the propositions that Book VI Proposition 9 depends on, it eventually comes back to Book I Proposition 4. $\endgroup$ Dec 29, 2013 at 17:13
  • $\begingroup$ This is a helpful source for those that are searching the Prepositions of the Elements, so I'll leave it here for now. $\endgroup$ Dec 29, 2013 at 21:17
  • $\begingroup$ I think you can prove it without using proposition 4, hilbert concluded that proposition 4 was not needed at all see en.wikipedia.org/wiki/Hilbert%27s_axioms but Euclid was not complete in his axiomatisation $\endgroup$
    – Willemien
    Jan 3, 2014 at 17:38
  • $\begingroup$ @Willemien Hilbert's axioms do include a side-angle-side postulate (III.6 in Wikipedia's numbering), but are you saying that Hilbert does not use that postulate in his proof that you can bisect a line segment? Do you have a page number or theorem number for Hilbert's proof in the Foundations of Geometry? $\endgroup$ Jan 3, 2014 at 23:56

1 Answer 1


The unavoidable step in the proof is the principle that opposite sides of a parallelogram are equal.

In Euclidean geometry this is proved using the congruence axiom that you want to avoid, but really it is a weaker principle from the subsystem of affine geometry that does not include the concepts of angle or rotation. In effect it is the Euclidean axiom restricted to superpositions that are parallel translations. (Maybe 180 degree rotations could be implicitly needed for some arguments, but they are avoidable for the midpoint proof by using a chain of several parallelograms, and they might be avoidable in general).

The two subsegments on either side of the midpoint are different geometric objects, and without some comparison principle no relation between them can be proved. There would be one segment divided into two parts but no other facts derivable. The minimal sufficient principle is the one about parallelograms.

  • $\begingroup$ How would you prove the result using only the principle that opposite sides of a parallelogram are equal? $\endgroup$ Jan 15, 2014 at 1:44
  • $\begingroup$ Take a parallelogram with the segment as one of its diagonals, and draw lines parallel to the diagonals through all four vertices. In the resulting figure the principle can be applied several times to get a chain of equalities that connect the two sub-segments. $\endgroup$
    – zyx
    Jan 15, 2014 at 1:51
  • $\begingroup$ Could you spell your proof out, preferably with a picture? I'm having trouble visualizing this. $\endgroup$ Jan 17, 2014 at 17:39
  • $\begingroup$ Take a parallelogram $P$ and the 4 midpoints of its sides, which are vertices of a second parallelogram $P'$. This is to be set up so that one of the diagonals of $P'$ is the segment to be bisected (pick any point not collinear with the segment as a third vertex of $P'$, and complete the diagram as described using parallel lines). In the figure consisting of $P$, $P'$ and the diagonals of $P'$, several applications of the equality principle for parallel segments produce a chain of equalities between the left and right half of the bisected segment. $\endgroup$
    – zyx
    Jan 19, 2014 at 5:55
  • $\begingroup$ Note that in the preceding comment, the word "midpoint" is used only to make it definite what the diagram will look like. To construct the diagram, from segment to $P'$ to $P$, does not require the ability to construct midpoints. $\endgroup$
    – zyx
    Jan 19, 2014 at 9:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .