Also called the law of cosines, the Al-Kashi's theorem is a generalization of the Pythagorean theorem for any triangle. It is expressed as follows:
In a context of an ordinary triangle \(\{a, b, c\}\), which every angle \(\alpha, \beta, \gamma \) in front of its correspondant side, such as:
$$ \left \{ \begin{gather*} \alpha \enspace opposed \enspace to \enspace a \\ \beta \enspace opposed \enspace to \enspace b \\ \gamma \enspace opposed \enspace to \enspace c \end{gather*} \right \} $$
and such as the following figure:
We do have the three following relations:
$$ a^2 = b^2 + c^2 - 2bc.cos(\alpha) \qquad (Al-Kashi) $$
$$ b^2 = a^2 + c^2 - 2ac.cos(\beta) \qquad (Al-Kashi^*) $$
$$ c^2 = a^2 + b^2 - 2ab.cos(\gamma) \qquad (Al-Kashi^{**}) $$
In all these demonstrations, we will demonstrate the theorem for only one of the three sides. Afterward the two other demonstrations are trivial.
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,
$$ a^2 = b^2 + c^2 - 2bc.cos(\alpha) \qquad (Al-Kashi) $$
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:
$$ a^2 = b^2 + c^2 - 2bc.cos(\alpha) \qquad (Al-Kashi) $$
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:
$$ \forall \vec{u},$$
$$ \vec{u}.\vec{u} = || \vec{u} ||^2$$
So,
$$ \vec{a}.\vec{a} = || \vec{a} ||^2$$
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:
$$ \forall (\vec{u}, \vec{v}), $$
$$ (\vec{u} - \vec{v})^2 = || \vec{u} ||^2 - 2 \vec{u}.\vec{v} + || \vec{v} ||^2 $$
So in our case,
$$ || \vec{b} ||^2 - 2 \vec{b}.\vec{c} + || \vec{c} ||^2 = || \vec{a} ||^2$$
$$ b^2 - 2 bc \ cos(\vec{b}, \vec{c}) + c^2 = a^2 $$
$$ a^2 = b^2 + c^2 - 2bc.cos(\alpha) \qquad (Al-Kashi) $$
We demonstrated the first formula of the theorem:
$$ a^2 = b^2 + c^2 -2bc.cos(\alpha) \qquad (Al-Kashi)$$
Repeating the same process again on the other sides of the triangle we can retrieve two other expressions.
$$ b^2 = a^2 + c^2 - 2ac.cos(\beta) \qquad (Al-Kashi^*) $$
$$ c^2 = a^2 + c^2 - 2ab.cos(\gamma) \qquad (Al-Kashi^{**}) $$
Go to the top of the page
