Legendre’s theorem on angles of triangle

Adrien-Marie Legendre has proved some theorems concerning the sum of the angles of triangle.  Here we give one of them, being the inverse of the theorem in the entry “sum of angles of triangle in Euclidean geometry”.

Theorem.  If the sum of the interior angles of every triangle equals straight angle, then the parallel postulate is true, i.e., in the plane determined by a line and a point outwards it there is exactly one line through the point which does not intersect the line.

Proof.  We consider a line $a$ and a point $B$ not belonging to $a$.  Let $BA$ be the normal line of $a$(with $A\in a$) and $b$ be the normal line of $BA$ through the point $B$.  By the supposition of the theorem, $b$ does not intersect $a$.

We will show that in the plane determined by the line $a$ and the point $B$, there are through $B$ no other lines than $b$ not intersecting the line $a$.  For this purpose, we choose through $B$ a line $b^{\prime }$ which differs from $b$; let the line $b^{\prime }$ form with $BA$ an acute angle $\beta$.

We determine on the line $a$ a point $A_1$ such that  $A{A}_1=AB$.  By the supposition of the theorem, in the isosceles right triangle $BA{A}_1$ we have

$${\alpha }_1=:\mathrm{\angle }A{A}_{1}B=\frac{\pi }{4}=\frac{\pi }{2^2}.$$

Next we determine on $a$ a second point $A_2$ such that $A_1{A}_2=A_{1}B$.  By the supposition of the theorem, in the isosceles triangle $B{A}_1{A}_2$ we have

$${\alpha }_2=:\mathrm{\angle }A{A}_{2}B=\frac{{\alpha }_1}{2}=\frac{\pi }{2^3}.$$

We continue similarly by forming isosceles triangles using the points $A_3$, $A_3$, $\mathrm{\dots }$, $A_n$ of the line $a$ such that

$$A_2{A}_3=B{A}_2,A_3{A}_4=B{A}_3,\mathrm{\dots },A_{n-1}{A}_n=B{A}_{n-1}.$$

Then the acute angles being formed beside the points are

$${\alpha }_3=\frac{\pi }{2^4},{\alpha }_4=\frac{\pi }{2^5},\mathrm{\dots },{\alpha }_n=\frac{\pi }{2^{n+1}}.$$

They form a geometric sequence with the common ratio $r=\frac{1}{2}$.  When $n$ is sufficiently great, the member ${\alpha }_n$ is less than any given positive angle.  As we have so much triangles $B{A}_{n-1}{A}_n$ that ,  then

$$\mathrm{\angle }AB{A}_n=\frac{\pi }{2}-{\alpha }_n>\beta .$$

Then the line $b^{\prime }$ falls after penetrating $B$ into the inner territory of the triangle $AB{A}_n$.  Thereafter it must leave from there and thus intersect the side $A{A}_n$ of this triangle.  Accordingly, $b^{\prime }$ intersects the line $a$.

The above reasoning is possible for each line  $b^{\prime }\ne b$  through $B$.  Consequently, the parallel axiom is in force.