The Pythagorean theorem and its reciprocal

The Pythagorean theorem

Let be a triangle \(\{a, b, c\}\), right-angled between \( a\) and \( b\) such as the following figure:

1. The classical demonstration

   To proove the truthfulness of the theorem, we projected a height \( h_c\) on the hypotenuse \( c\).

   

   We know that the sum of the angles of a triangle is equal to \(\pi \enspace (180°)\) .

   In the main triangle formed by \(\{a, b, c\}\), we notice that \(\alpha + \beta + \frac{\pi}{2} = \pi\).

   This general relation will allow us to deduce other angles.

   Now, in the triangle formed by \(\{m, \; h_c, \; a\}\), we do have a right angle and the angle \(\beta\). Therefore, the third angle is \(\alpha\).

   Finally, in the last triangle formed by \(\{h_c, \; n, \; b\}\), we do have a right angle and the angle \(\alpha\). Thus, the third angle is \(\beta\).

   We added them to the following figure:

   

   A property of the similar triangles tells us that when two triangles have two-by-two the same angles, they are similar , and will then have two-by-two similar sides forming the same ratio.

   In this case, pn has the following relations:

   $$ \frac{a}{c} = \frac{m}{a} = \frac{h_c}{b} $$

   $$ \frac{a}{c} = \frac{m}{a} \qquad (1)$$

   $$ \frac{b}{c} = \frac{n}{b} = \frac{h_c}{a} $$

   $$ \frac{b}{c} = \frac{n}{b} \qquad (2)$$

   Thanks to both expressions \((1)\) and \((2)\), we then have:

   $$ \Biggl \{ \begin{gather*} a^2 = cm \qquad (3) \\ b^2 = cn \qquad (4) \end{gather*} $$

   Now, additionning \((3) \) and \((4)\), we do obtain:

   $$ a^2 + b^2 = cm + cn$$
   $$ a^2 + b^2 = c(m + n) $$

   But \( (m + n = c) \), so as a result,

   $$ a \perp b \Longrightarrow a^2 + b^2 = c^2 \qquad \bigl(\text{Pythagorean theorem} \bigr) $$
2. Einstein's excellent intuition

   As in the previous demonstration, Einstein had noted that when the height projected onto the hypotenuse is drawn, it forms three similar triangles .

   

   Now, such right-angled similar triangles , the areas of each triangle were proportional to the square of their respective hypotenuse.

   Indeed, if we start from any triangle \(\{a ,\ b, \ c\}\) which is rectangle between \( a\) and \( b\), if we do apply a proportionnality coefficient \(k\) to obtain a second triangle \(\{a', \; b', \; c'\}\), then:

   $$ \left \{ \begin{gather*} a' = ka \\ b' = kb \\ c' = kc \end{gather*} \right \} $$

   And the respective areas of the two triangles also undergo the transformation:

   $$ A = \frac{ab}{2} \longmapsto A' = \frac{ka \times kb}{2} $$
   $$ A' = k^2 \times \frac{ab}{2} $$
   $$ A' = k^2 \times A $$
   
   

   Since the two internal triangles are simply reductions of the large triangle, we can define the following two coefficients:

   $$ \left \{ \begin{gather*} \text{Little triangle} : k_1 = \frac{a}{c} \\ \\ \text{Middle triangle} : k_2 = \frac{b}{c} \end{gather*} \right \} $$

   As well as their respective areas depending on \(A\):

   $$ \left \{ \begin{gather*} \text{Little triangle} : A_1 = k_1 \times A \\ \\ \text{Middle triangle} : A_2 = k_2 \times A \end{gather*} \right \} $$

   By substituting, we obtain:

   $$ \left \{ \begin{gather*} \text{Little triangle} : A_1 = \left( \frac{a}{c} \right)^2 \times A \\ \\ \text{Middle triangle} : A_2 = \left( \frac{b}{c} \right)^2 \times A \end{gather*} \right \} $$

   And \since areas add up:

   $$ A = A_1 + A_2 $$
   $$ A = \left( \frac{a}{c} \right)^2 \times A + \left( \frac{b}{c} \right)^2 \times A $$
   $$ A = A \left[ \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 \right] $$

   Finally, we can deduce that:

   $$ \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 = 1 $$
   $$ \frac{a^2 + b^2}{c^2} = 1 $$
   
   

   And finally, that,

   $$ a \perp b \Longrightarrow a^2 + b^2 = c^2 \qquad \bigl(\text{Pythagorean theorem} \bigr) $$
3. Garfield's demonstration

   A demonstration with a more pronounced geometric aspect.

   If we draw two identical right triangles \(\{a ,\ b, \ c\}\), one flattened on \(b\) and the other flattened on \(a\) as shown in the following figure:

   

   So, the two hypotenuse \(c\) are adjacents.

   Furthermore, \since the sum of the angles in a triangle is constant and the third angle is a right angle, we have:

   $$ \alpha + \beta = \frac{\pi}{2} $$

   However, a flat angle is worth \(\pi\), e angle formed by the two hypotenuses \(c\) are necessarily worth \(\frac{\pi}{2}\). Therefore, a right angle is added to the figure:

   

   We now have a trapezoid whose area we can calculate:

   $$ A = (a + b) \times \frac{(a + b)}{2} $$
   $$ A = \frac{(a + b)^2}{2} $$

   Developping the expression we do obtain this:

   $$ A = \frac{a^2 + 2ab + b^2}{2} $$
   $$ A = \frac{a^2 + b^2}{2} + 2 \times \frac{ab}{2} $$

   The areas of the two right-angle triangles \(\{a ,\ b, \ c\}\) are identified:

   $$ \left \{ \begin{gather*} 2 \times \frac{ab}{2} : \text{Area of two right-angle triangles } \{a ,\ b, \ c\} \end{gather*} \right \} $$

   And by addition, the other is necessarily worth:

   $$ \left \{ \begin{gather*} \frac{a^2 + b^2}{2} : \text{Area of the half square of side } c \end{gather*} \right \} $$

   But this area of ​​the half-square is also, by definition, equal to \(\frac{c^2}{2}\), thus:

   $$ \frac{a^2 + b^2}{2} = \frac{c^2}{2} $$
   
   

   And finally, we find it all,

   $$ a \perp b \Longrightarrow a^2 + b^2 = c^2 \qquad \bigl(\text{Pythagorean theorem} \bigr) $$

The Pythagorean theorem reciprocal

To proove now the truthfulness of the theorem reciprocal, we start from a triangle, a priori right-angled, but let start from the only hypothesis that:

1. By calculating the angle \(\gamma\), a priori right-angled

   Let us project the height \( h_c \) intersecting the side \( c \) at right-angle, and splitting the angle \( \gamma \) in two differents angles \( \gamma_a \) and \( \gamma_b \):

   

   If \(\gamma\) is a right angle, then \(\cos(\gamma) = 0\).

   We know from the trigonometric addition formulas that:

   $$ \forall (\alpha, \beta) \in \hspace{0.04em} \mathbb{R}^2, $$

   $$ \cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) $$

   So in our case,

   $$ \cos(\gamma) = \cos(\gamma_a + \gamma_b) = \cos(\gamma_a) \cos(\gamma_b) - \sin(\gamma_a) \sin(\gamma_b) $$
   $$ \cos(\gamma) = \frac{h_c}{a} . \frac{h_c}{b} - \frac{m}{a} . \frac{n}{b} $$
   $$ \cos(\gamma) = \frac{h_c^2 - mn}{ab} \qquad (5) $$

   With the following equations which can be noticed on the above figure,

   $$ \left \{ \begin{gather*} a^2 = m^2 + h_c^2 \\ b^2 = n^2 + h_c^2 \\ c^2 = (m+n)^2 = m^2 + 2mn + n^2 \end{gather*} \right \} $$

   we see that our hypothesis:

   $$ a^2 + b^2 = c^2 $$

   becomes:

   $$ \underbrace{ m^2 + h_c^2 } _{ a^2 } \enspace + \enspace \underbrace{ n^2 + h_c^2} _{ b^2 } \enspace = \enspace \underbrace{ m^2 + 2mn + n^2 } _{ c^2 }$$
   $$ 2h_c^2 + m^2 + n^2 = 2mn + m^2 + n^2 $$
   $$ h_c^2 = mn $$

   And then,

   $$ h_c^2 - mn = 0 \qquad (6) $$

   Now, injecting \( (6) \) into \( (5) \) we do obtain:

   $$ \cos(\gamma) = 0 \Longleftrightarrow \Biggl \{ \gamma = \frac{\pi}{2} \ ou \ \gamma = -\frac{\pi}{2} \Biggr \} $$

   That definitely means that the angle \( \gamma \) is a right angle, and as a consequence of it the triangle \(\{a, b, c\}\) is right-angled between \(a\) and \(b\).

   $$ a^2 + b^2 = c^2 \Longrightarrow a \perp b \qquad \bigl(\text{Pythagorean theorem (reciprocal)} \bigr) $$

2. By comparing areas

   Let us project again the height \( h_c \) intersecting the side \( c \) at right angle.

   

   We know that the triangle area is worth:

   $$ S_{triangle} = \frac{\text{base} \times \text{height}}{2} \qquad (7) $$

   But, the area of such triangle is worth:

   $$ S_{triangle} = \frac{1}{2} \sin(\gamma) \times a b \qquad (8) $$

   And combining \((7)\) and \((8)\) :

   $$ \frac{c.h_c}{2} = \frac{\sin(\gamma) \times a b}{2} \Longleftrightarrow c.h_c = \sin(\gamma) \times a b $$

   We now want to show that \(\sin(\gamma) = 1\) such as the triangle is definitely a right-angled triangle.

   To do this, let us start from our initial hypothesis:

   $$ a^2 + b^2 = c^2 $$

   And,

   $$ a^2 + b^2 = (m + n)^2 \qquad (9) $$

   We now know from the Pythagorean theorem that in the inner triangle \(\{a, m, h_c\}\):

   $$ a^2 = m^2 + h_c^2 $$

   So that:

   $$ m^2 = a^2 - h_c^2 $$

   $$ m = \sqrt{ a^2 - h_c^2} \qquad (10) $$

   As well as the other inner triangle:

   $$ n = \sqrt{ b^2 - h_c^2} \qquad (11) $$

   Now injecting \( (10) \) and \( (11) \) into \( (9) \), we do have:

   $$ a^2 + b^2 = \left(\sqrt{ a^2 - h_c^2 } + \sqrt{ b^2 - h_c^2} \right)^2 $$

   Distributing it, we obtain:

   $$ a^2 + b^2 = a^2 - h_c^2 + 2\sqrt{ (a^2 - h_c^2)( b^2 - h_c^2)} + b^2 - h_c^2 $$

   Let us remove the member \( (a^2 + b^2) \) which is present on both sides oh the equation:

   $$ 0 = -2h_c^2 + 2\sqrt{ (a^2 - h_c^2)( b^2 - h_c^2)} $$
   $$ 2h_c^2 = 2\sqrt{a^2b^2 -a^2h_c^2 - b^2h_c^2 + h_c^4} $$
   $$ h_c^2 = \sqrt{a^2b^2 -a^2h_c^2 - b^2h_c^2 + h_c^4} $$

   Then, let us apply the square to get rid of the square root:

   $$ \left(h_c^2 \right)^2 = \left(\sqrt{a^2b^2 -a^2h_c^2 - b^2h_c^2 + h_c^4}\right)^2 $$
   $$ h_c^4 = a^2b^2 -a^2h_c^2 - b^2h_c^2 + h_c^4 $$

   We can remove \( h_c^4 \) present on both sides.

   $$ a^2b^2 = a^2h_c^2 + b^2h_c^2 $$

   We factorize it:

   $$ a^2b^2 = h_c^2 (a^2 + b^2) $$

   But, our initial hypothesis was that:

   $$ a^2 + b^2 = c^2 $$

   Thus,

   $$ a^2 \ b^2 = h_c^2 \ c^2 $$
   $$ \sqrt{a^2 \ b^2} = \sqrt{h_c^2 \ c^2 } $$
   $$ c \ h_c = a \ b $$

   However, we previous had this result :

   $$c \ h_c = \sin(\gamma) \times a \ b $$

   That automatically implies taht \(\sin(\gamma)= 1\), and that the angle \(\gamma\) is a right angle.

   We definitely showed that the triangle \(\{a, b, c\}\) is right-angled between \(a \) and \( b \). Hence:

   $$ a^2 + b^2 = c^2 \Longrightarrow a \perp b \qquad \bigl(\text{Pythagorean theorem (reciprocal)} \bigr) $$

The Pythagorean theorem equivalence

Two implications makes an equivalence.

Thus, having our two implications \((I_1)\) et \((I_2)\):

We can gather them to build the following equivalence: