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Recall that the midline theorem in a Euclidean plane says that the segment between the midpoints of a triangle is parallel to the third side and half as long.

I wonder if we exclude the axiom of parallels, that is, consider a Hilbert plane where the axiom of parallels is not satisfied, is it true that the above mentioned segment is strictly less than a half of the third side? I cannot come up with a proof.

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    What would be the definition of a segment? – 2017-02-27
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    The set of all points between P and Q (if we call the midpoints of the first two sides P and Q). – 2017-02-27
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    What does "between" mean, say, in the case of a Hilbert space? – 2017-02-27
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    It is an undefined notion. – 2017-02-27

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Preliminary note

The Statement in the OP cannot be proved without the hyperbolic version of the axiom of parallels. That is, we will assume that in the plane to a straight line through any point not on that straight line at least two parallels can be constructed. (The rest of the Hilbert axioms are assumed to be true.)

First, let's recall a trivial absolute statement (without proof).

enter image description here

If $AB=A'B'$ and $AC=A'C'$ and $\angle CAB \ < \ \angle C'A'B'$ then $CB

Second, in hyperbolic geometry the sum of the angles of a triangle is less than $180^{\circ}$

Third. Consider a quadrangle.

enter image description here

Let $AD=BC$ and let $\angle DAB+\angle ABC=180^{\circ}$ then $AB

Proof

Taking the the triangle $\triangle DAB$ $$\angle DAB+\angle ABD+\angle ADB\ <\ 180^{\circ}=\angle DAB+\angle ABC=\angle DAB+\angle ABD+\angle DBC.$$

Thus

$$\angle ADB\ <\ \angle DBC.$$

Applying the First statement in the case of the triangles $\triangle DAB$ and $\triangle DBC$ we conclude that

$$AB < DC.$$

Fourth The OP's theorem

Consider the triangle $\triangle ABC$ below. Let $F$ and $E$ be the midpoints of $BA$ and $CA$. We have to prove that $FE<\frac12 BC$ or, in other words $2FE

enter image description here

Extend the midline $FE$ so that $EF=EH$. Now we have a quadrangle $BFHC$. It is easy to see that $\triangle FAE$ and $\triangle EHC$ are congruent. Thus the two sides of the rectangle $BFHC$ equal: $BF=CH$. Also, the angles $\angle AFE=\angle EHC$. We can say then that $180^{\circ}=\angle BFE+\angle AFE=\angle BFE+\angle EHC=180^{\circ}.$

The assumptions of the Third assertion are met. Thus we can tell that

$$2FE