I have been looking around for a proof by contradiction on $AAS$ congruence in neutral geometry, but can not find any sources on it. Does anyone know how the proof by contradiction goes for angle angle side congruence in neutral geometry?
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$\begingroup$Suppose we have a triangle with one side length $c$ and two angles $\alpha$ and $\gamma$, where $\gamma$ is the angle opposite $c$. We must show that this triangle is unique up to congruence.
Let us construct this triangle.
We start by drawing segment $AB$ of length $c$. Then draw a ray starting at $A$ which forms angle $\alpha$ with segment $AB$. So far everything is unique up to congruence.
Now to complete the proof, we must show that there is at most one point $C$ on the above ray such that angle BCA is equal to $\gamma$. If there were 2 such points $C$ and $C'$, with angles $BCA$ and $BC'A$ both equal to $\gamma$, then angles of triangle $BCC'$ add up to more than $180^{\circ}$. This is a contradiction in neutral geometry.
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