Saccheri-Legendre theorem

Question

In neutral geometry , you know that :

" The sum of the angles in a triangle is at most 180 degree. "

I seek any proof of this theorem but I can't find. Would you help me?

Answer 1

Consider the line $e$ and the triangle $A_1A_2B_1$ such that $A_1A_2$ lie on $e$ as shown in the following figure.

Now, copy this triangle on $e$ many times such that $A_1A_2B_1$ is congruent with $A_2A_3B_2$; $A_2A_3B_3$ is congruent with $A_3A_4B_3$, and so on. (Note that the union of the segments $B_i,B_{i+1}$ don't necessarily form a straight line. In the history of geometry it was one of the major mistakes that some took it granted that $B_1,B_2,...,B_n$ lie on a straight line. If this is the case then we have the Euclidean geometry.)

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Assume that $\alpha+\beta+\gamma>\pi$. Obviously $\beta+\gamma'+\alpha=\pi$. Subtract the second equation from the first one and get that $\gamma-\gamma'>0$. That is $\gamma>\gamma'$.

If $\gamma>\gamma'$ then $A_1A_2>B_1B_2$, therefore there exists an $\varepsilon>0$ such that $A_1A_2-B_1B_2>\varepsilon$. The corresponding triangles are congruent here and the inequality above holds for all the pairs of segments $A_iA_{i+1}-B_iB_{i+1}$. As a result, if $n$ is large enough the sum of the differences can be greater than any given length. Choose $n$ such that

$$(n-1)A_iA_{i+1}-\sum_{i=1}^{n-1}B_iB_{i+1}>A_1B_1+A_nB_n$$

or

$$(n-1)A_iA_{i+1}>A_1B_1+A_nB_n+\sum_{i=1}^{n-1}B_iB_{i+1}.\tag 1$$

Notice that

$$(n-1)A_iA_{i+1}=A_1A_n.$$

and because of $(1)$

$$A_1A_n>\sum_{i=1}^{n-1}B_iB_{i+1}+A_1B_1+A_nB_n.$$

So, if $\alpha+\beta+\gamma>\pi$ then the length of $A_1A_n$ is greater than the length $A_1B_1+\sum_{i=1}^{n-1}B_iB_{i+1}+B_nA_n$ . This is impossible because $A_1A_n$ connects $A_1$ and $A_n$ along a straight line while the path $A_1B_1B_2...B_nA_n$ connects $A_1$ and $A_n$ along a path different from the straight $A_1A_n$ as shown below:

Therefore, assuming that the sum of the angles of a triangle is greater than $\pi$ results in a contradiction.