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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

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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?.

Any comments about this will be appreciated.

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I managed to prove this theorem in hyperbolic geometry, using the hyperbolic cosine law:

$$ \cosh a=\cosh b\cosh c -\sinh b\sinh c\cos\alpha, $$ 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 $a=b$.

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