The Pythagorean theorem and its reciprocal

Demonstration

The Pythagorean theorem

1. The classical demonstration

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



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 are identified:

$$ \left \{ \begin{gather*} 2 \times \frac{ab}{2} : \text{Aire des deux triangles rectangles } \{a, \; b, \; c\} \end{gather*} \right \} $$

And by addition, the other is necessarily worth:

$$ \left \{ \begin{gather*} \frac{a^2 + b^2}{2} : \text{Aire du demi-carré de côté } c \end{gather*} \right \} $$

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 } _\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 \(b\).

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

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:

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

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:

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

The Pythagorean theorem equivalence

Two implications makes an equivalence.

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

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

We can gather them to build the following equivalence:

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:

$$\forall (A, B) \in \hspace{0.04em} (O, \vec{x}, \vec{y})^2, $$
$$AB = \sqrt{ (x_b - x_a)^2 + (y_b - y_a)^2} $$

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:

$$\forall (A, B) \in \hspace{0.04em} (O, \vec{x}, \vec{y}, \vec{z})^2, $$
$$AB =\sqrt{(x_b - x_a)^2 + (y_b - y_a)^2 + (z_b - z_a)^2 }$$