# If two medians are congruent… is the triangle isosceles in a Hilbert plane?

If $$ABC$$ is a triangle for which two medians are congruent... is it true that the triangle $$ABC$$ is isosceles in a general Hilbert plane? I am having a little bit of trouble trying to prove this, if it is true. Here is my try, in a Hilbert Plane with the parallelism axiom:

First, let $$CD,\,BE$$ be the medians of the sides $$AC$$ and $$AB$$, respectively. Let $$M$$ be the point where these medians cut (that exists, as an easy application of Pasch's axiom).

By hypothesis, $$AD=BD$$, $$AE=CE$$, and $$BE=CD$$ (hence $$AB=2AD=2BD,\,AC=2AE=2CE)$$. Assuming the parallelism axiom, we have by Thales theorem (VI 2), that, since:

$$\frac{AB}{AD}=\frac{AC}{AE}$$

$$DE$$ is parallel to $$BC$$, and the triangles $$ADE$$ and $$ABC$$ are similar. In particular, $$BC=2DE$$. Applying (I29) twice, we can conclude that the triangles $$BCM$$ and $$EDM$$ have all the same angles, so by (VI 4), we conclude that these are similar triangles. But then:

$$\frac{BC}{ED}=\frac{CM}{DM}$$

And the left hand side equals $$2$$, so we conclude that $$CM=2DM$$, or in other words, $$-$$ since $$CD=CM+MD\,-$$, that $$CD=\frac{3}{2}CM$$.

On the other hand, we also have that:

$$\frac{BC}{ED}=\frac{BM}{EM}$$

And since the left hand side is equal to $$2$$, we also obtain that $$BM=2EM$$, or simply, that $$BE=\frac{3}{2}BM$$

But by hypothesis, $$BE=\frac{3}{2}BM=\frac{3}{2}CM=CD$$, therefore $$BM=CM$$. Applying (I6) to the isosceles triangle $$MBC$$ we conclude that $$\angle MBC=\angle MCB$$. And then:

$$\begin{cases} \ BE=CD\,\text{ (by hypothesis)} \\ \ \angle EBC=\angle DCB\,\text{ (since we just proved that }\angle MBC=\angle MCB) \\ \ CB=BC\,\text{ (C2)} \\ \end{cases}$$

So by (C6) we conclude that the triangles $$EBC$$ and $$DCB$$ are congruent. In particular, $$CE=BD$$. Since $$AB=2BD$$ and $$AC=2CE$$, we conclude that $$AB=AC$$, and then the triangle $$ABC$$ is isosceles

This proof relied heavily in the theory of similar triangles that can only be developed assuming the parallelism axiom. However, can this be proved in a more general context, avoiding the use of the parallelism axiom, or any of its consequences? Can there be a non-euclidean model of the Hilbert plane in which this statement is false?.

$$\cosh a=\cosh b\cosh c -\sinh b\sinh c\cos\alpha,$$ where $$a$$, $$b$$, $$c$$ are the sides of a triangle and $$\alpha$$ the angle opposite to $$a$$.
If $$m$$ is the length of the median joining the midpoint of side $$b$$ with the opposite vertex, then we have: $$\cosh m=\cosh{b\over2}\cosh c -\sinh {b\over2}\sinh c\cos\alpha= {\cosh c+\cosh a\over2\cosh{b\over2}},$$ where I substituted $$\cos\alpha$$ from the previous equation and used the identity: $$\cosh b=2\cosh{b\over2}-1$$.
In the same way, if $$n$$ is the length of the median joining the midpoint of side $$a$$ with the opposite vertex, we obtain: $$\cosh n={\cosh c+\cosh b\over2\cosh{a\over2}}.$$ If $$m=n$$ we then get the equality $$\cosh{b\over2}{(\cosh c+\cosh b)}=\cosh{a\over2}{(\cosh c+\cosh a)},$$ which can be rewritten as $$\left(\cosh{b\over2}-\cosh{a\over2}\right) \left[\cosh c-1+2\left(\cosh^2{b\over2}+\cosh^2{a\over2} +\cosh{b\over2}\cosh{a\over2} \right)\right]=0.$$ The expression between square brackets is positive, hence it must be $$\cosh{b\over2}=\cosh{a\over2}$$, that is $$a=b$$.