The Pythagorean theorem and its reciprocal

The Pythagorean theorem

The Pythagorean theorem tells us that:

In a right-angle triangle, the square of the hypotenuse is equal to the sum of the square on the sides of the right angle.

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

We do have the following equation:

$$ a \perp b \Longrightarrow a^2 + b^2 = c^2 \qquad \bigl(Pythagore \enspace(theorem) \bigr) $$

The Pythagorean theorem reciprocal

Its reciprocal tells us the contrary:

In any triangle \(\{a, b, c\}\), where \( c\) is the longest side:

$$ a^2 + b^2 = c^2 \Longrightarrow a \perp b \qquad \bigl(Pythagore \enspace(reciprocal) \bigr) $$

The Pythagorean theorem equivalence

Both previous implications then form the following equivalence:

$$ a \perp b \Longleftrightarrow a^2 + b^2 = c^2 \qquad \bigl(Pythagore \enspace(equivalence) \bigr) $$

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Demonstration

The Pythagorean theorem

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

We have the following trigonometric relations:

$$ cos(\beta) = \frac{a}{c} = \frac{m}{a} \qquad (1)$$ $$ cos(\alpha) =\frac{b}{c} = \frac{n}{b} \qquad (2)$$

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

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

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

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

$$ a \perp b \Longrightarrow a^2 + b^2 = c^2 \qquad \bigl(Pythagore \enspace(theorem) \bigr) $$

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:

$$ \forall (\alpha, \beta) \in \hspace{0.05em} \mathbb{R}^2, $$

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

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,

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

we see that our hypothesis:

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

becomes:

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

$$ a^2 + b^2 = c^2 \Longrightarrow a \perp b \qquad \bigl(Pythagore \enspace(reciprocal) \bigr) $$

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:

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

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:

$$ a^2 + b^2 = c^2 \Longrightarrow a \perp b \qquad \bigl(Pythagore \enspace(reciprocal) \bigr) $$

The Pythagorean theorem equivalence

Two implications makes an equivalence.

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

$$ \Biggl \{ \begin{align*} a \perp b \Longrightarrow a^2 + b^2 = c^2 \qquad (I_1)\\ a^2 + b^2 = c^2 \Longrightarrow a \perp b \qquad (I_2) \end{align*} $$

We can gather them to build the following equivalence:

$$ a \perp b \Longleftrightarrow a^2 + b^2 = c^2 \qquad \bigl(Pythagore \enspace(equivalence) \bigr) $$

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Example

The Pythagorean theorem allows us to measure lengths on both plane and space.

1. Calculate a length in the plane

We start from a plane \((\vec{x}, \ \vec{y}) \) in which it exists two points \( A(x_a, \ y_a )\) and \(B(x_b, \ y_b )\).



Joining on the abscissa axis \( x_a \) and \( x_b\), as well as on the ordinate axis \( y_b \) and \( y_a\), we obtain a third point \( C\) and as a consequence of it a right-angled triangle \(ABC \), right-angled in \(C \).



Thus, we can apply the Pythagorean theorem on it:

$$AB^2 = AC^2 + BC^2$$
$$AB^2 = (x_b - x_a)^2 + (y_b - y_a)^2 $$
$$AB = \sqrt{ (x_b - x_a)^2 + (y_b - y_a)^2} $$

Thus, the distance \( AB\) in a two-dimensional space is worth:

$$\forall (A, B) \in \hspace{0.05em} (O, \vec{x}, \vec{y})^2, $$

$$AB = \sqrt{ (x_b - x_a)^2 + (y_b - y_a)^2} $$

2. Calculate a length in space

We want now to calculate a length \(AB \) in space.



We previously found out that the length \(AC \), on this new figure, is worth:

$$AC = \sqrt{ (x_b - x_a)^2 + (y_b - y_a)^2} $$


Consequently, we do apply again the Pythagorean theorem on the triangle \(ABC \):

$$AB^2 = AC^2 + BC^2$$
$$AB^2 = (x_b - x_a)^2 + (y_b - y_a)^2 + (z_b - z_a)^2 $$
$$AB = \sqrt{(x_b - x_a)^2 + (y_b - y_a)^2 + (z_b - z_a)^2 } $$

Thus, the distance \( AB\) in a three-dimensional space is worth:

$$\forall (A, B) \in \hspace{0.05em} (O, \vec{x}, \vec{y}, \vec{z})^2, $$

$$AB =\sqrt{(x_b - x_a)^2 + (y_b - y_a)^2 + (z_b - z_a)^2 }$$