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:
$$ a \perp b \Longleftrightarrow a^2 + b^2 = c^2 \qquad \bigl(Pythagore \enspace(equivalence) \bigr) $$
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:
$$ a^2 + b^2 = c.m + c.n$$
$$ a^2 + b^2 = c.(m + n) $$
But \( (m+n= c) \), so as a result,
$$ a \perp b \Longrightarrow a^2 + b^2 = c^2 \qquad \bigl(Pythagore \enspace(theorem) \bigr) $$
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:
$$ a^2 + b^2 = c^2 $$
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 \(b\).
$$ a^2 + b^2 = c^2 \Longrightarrow a \perp b \qquad \bigl(Pythagore \enspace(reciprocal) \bigr) $$
Let us project again the height \( h_c \) intersecting the side \( c \) at right angle.
We know that the triangle area is worth:
$$ S_{triangle} = \frac{base.hauteur}{2} $$
In our case, that would be:
$$ S_{triangle} = \frac{c.h_c}{2} \qquad (7) $$
Moreover, if there was a right angle between the sides \( a \) and \( b \), then \( a \) (as well as \( b \)) would be a height of the triangle \(\{a, b, c\}\), and as a consequence we would also have:
$$ S_{triangle} = \frac{a.b}{2} \qquad (8) $$
Les deux surfaces du triangle \( (7) \) et \( (8) \) étant équivalentes, on obtiendra l'équivalence :
$$ \frac{c.h_c}{2} = \frac{a.b}{2} \Longleftrightarrow c.h_c = a.b $$
That is what we will show.
TLet 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,
$$ a^2 \ b^2 = h_c^2 \ c^2 $$
$$ \sqrt{a^2 \ b^2} = \sqrt{h_c^2 \ c^2 } $$
And finally,
$$ c \ h_c = a \ b $$
That shows that \( a \) is definitely a height of the triangle \(\{a, b, c\}\), and as a consequence of it this triangle is right-angled between the sides \(a \) and \( b \).
$$ a^2 + b^2 = c^2 \Longrightarrow a \perp b \qquad \bigl(Pythagore \enspace(reciprocal) \bigr) $$
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:
$$ a \perp b \Longleftrightarrow a^2 + b^2 = c^2 \qquad \bigl(Pythagore \enspace(equivalence) \bigr) $$
The Pythagorean theorem allows us to measure lengths on both plane and space.
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:
$$\forall (A, B) \in \hspace{0.05em} (O, \vec{x}, \vec{y})^2, $$
$$AB = \sqrt{ (x_b - x_a)^2 + (y_b - y_a)^2} $$
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:
$$\forall (A, B) \in \hspace{0.05em} (O, \vec{x}, \vec{y}, \vec{z})^2, $$
$$AB =\sqrt{(x_b - x_a)^2 + (y_b - y_a)^2 + (z_b - z_a)^2 }$$
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