The geometric laws of the triangle

Law of \sines

To show it, let us project a height \( h_c \) upon the length \( c \), and such as the following figure:

Straightaway, the following relations come:

Dividing the equation \( (1) \) by \( a \), we do have:

In the same way, dividing \( (2) \) by \( b \):

We now notice that both right memebers of \( (3) \) and \( (4) \) are equals, it follows that:

By reproducing this operation on the two others lengths, we will have two new equations:

Equalities \( (5), (6), (7) \) having a common member from one to another, they are all equals.

And finally,

Al-Kashi's theorem

In all these demonstrations, we will demonstrate the theorem for only one of the three sides. Afterward the two other demonstrations are trivial.

1. With the Pythagorean theorem

   1. Case of an acute triangle

      To demonstrate the theorem, we have projected the height \( h_c \) on the side \( c \) to obtain the following figure:

      

      In the small triangle \(\{a, h_c, m\}\), the Pythagorean theorem gives us:

      $$ a^2 = m^2 + h_c^2 $$
      $$ a^2 = (c-n)^2 + h_c^2 $$
      $$ a^2 = c^2 - 2cn + n^2 + h_c^2 \qquad (1) $$

      Now, in the other triangle \(\{b, h_c, n\}\),

      $$ n^2 + h_c^2 = b^2 \qquad (2) $$

      But also,

      $$ \cos(\alpha) = \frac{n}{b} \Longleftrightarrow n = b.\cos(\alpha) \qquad (3) $$

      Injecting \((2)\) and \((3)\) into \((1)\), we do have now:

      $$ a^2 = c^2 - 2cb.\cos(\alpha) + b^2 $$

      And finally,

      $$\forall (a, b, c) \in \hspace{0.04em} \mathbb{R}^3, \enspace \forall \alpha \in \mathbb{R}, $$
      $$ a^2 = b^2 + c^2 - 2bc \cdot \cos(\alpha) \qquad (Al-Kashi) $$

   2. Case of an obtuse triangle

      

      In the same way, in the small triangle \(\{a, h_c, m\}\), the Pythagorean theorem gives us that:

      $$ a^2 = m^2 + h_c^2 $$
      $$ a^2 = (n-b)^2 + h_c^2 $$
      $$ a^2 = n^2 - 2nb + b^2 + h_c^2 \qquad (4) $$

      Now, in the other triangle \(\{h_c, n, c\}\),

      $$ h_c^2 + n^2 = c^2 \qquad (5) $$

      But also,

      $$ \cos(\alpha) = \frac{n}{c} \Longleftrightarrow n = c.\cos(\alpha) \qquad (6) $$

      Injecting \((5)\) and \((6)\) into \((4)\), we do have now:

      $$\forall (a, b, c) \in \hspace{0.04em} \mathbb{R}^3, \enspace \forall \alpha \in \mathbb{R}, $$
      $$ a^2 = b^2 + c^2 - 2bc \cdot \cos(\alpha) \qquad (Al-Kashi) $$

2. By the scalar product

   Considering the lengths \(a, b, c\) as three vectors, and such as the follwing figure:

   

   Then, we do have applying the Chasles relation :

   $$ \vec{a} = \vec{b} - \vec{c} \qquad (7) $$
   

   We know from the squared vector property that:

   $$ \forall \vec{u},$$

   $$ \vec{u}\cdot\vec{u} = {|| \vec{u} ||}^2$$

   So,

   $$ \vec{a}\cdot\vec{a} = {|| \vec{a} ||}^2$$

   Injecting \((7)\) into it, we now have:

   $$ (\vec{b} - \vec{c})^2 = {|| \vec{a} ||}^2$$

   Likewise, with the scalar product remarkable identities , we do have the following property:

   $$ \forall (\vec{u}, \vec{v}), $$

   $$ (\vec{u} - \vec{v})^2 = {|| \vec{u} ||}^2 - 2 \vec{u}\cdot\vec{v} + {|| \vec{v} ||}^2 $$

   So in our case,

   $$ {|| \vec{b} ||}^2 - 2 \vec{b}\cdot\vec{c} + {|| \vec{c} ||}^2 = {|| \vec{a} ||}^2$$
   $$ b^2 - 2 bc \ \cos(\vec{b}, \vec{c}) + c^2 = a^2 $$

   And finally,

   $$\forall (a, b, c) \in \hspace{0.04em} \mathbb{R}^3, \enspace \forall \alpha \in \mathbb{R}, $$
   $$ a^2 = b^2 + c^2 - 2bc \cdot \cos(\alpha) \qquad (Al-Kashi) $$

3. Conclusion

   We demonstrated the first formula of the theorem:

   $$ a^2 = b^2 + c^2 -2bc \cdot \cos(\alpha) \qquad (Al-Kashi)$$

   Repeating the same process again on the other sides of the triangle we can retrieve two other expressions.

   $$ b^2 = a^2 + c^2 - 2ac \cdot \cos(\beta) \qquad (Al-Kashi^*) $$
   $$ c^2 = a^2 + b^2 - 2ab \cdot \cos(\gamma) \qquad (Al-Kashi^{**}) $$

Area formula

To demonstrate this, let's project a height \( h_c \) along the length \( c \), and such as the following figure:

Immediately, the following relationships come:

As a result, it follows that:

Now, by injecting the values of \(h_c\) coming from \((1')\) and \((2')\) in that of \((\mathcal{A})\), we obtain the double equality:

Finally, by projecting one of the other two heights, we find the third equality, and finally:

Héron's formula

To demonstrate this formula, we project the height \( h_c \) as before along the length \( c \):

Thanks to Al-Kashi's theorem , we have the following relationship:

And then,

Now injecting the expression \((3)\) into the expression \((4)\), we do have this:

Arranging both parenthesis in order to obtain remarkable identities :

We now factorize to obtain the third remarkable identity :

We remove the parentheses:

So, by injecting the expression \((5)\) into the expression \((6)\), we do have:

Putting the elements in order, we now have:

At this stage, let us introduce the perimeter \(P\) of the triangle, then the expression becomes:

Then that of half-perimeter \(p\) :

We factorize all expressions in parentheses by \(2\):

And finally,

For any triplet of lengths \((a, b, c)\) of a triangle: