If you want to know if a triangle is an acute angled triangle, an obtuse angled triangle or a right angled triangle, and you're given only the slopes of the equations of its sides and don't want nor is requested to find the actual trigonometric values of its angles, the following theorem enable you to readily ascertain the triangle classification in the easyest way without further computations:
Let $m_1,m_2, m_3$ be the slopes of the cartesian equation of sides BC, CA, AB of a triangle whose internal angles are A, B, C. Let also $$P=(1+m_1m_2)(1+m_1m_3)(1+m_2m_3)$$ Then we may say that:
I.The triangle is a right angled triangle if and only if $P=0$
II.The triangle is an acute angled triangle if and only if $P<0$
III.The triangle is an obtuse angled triangle if and only if $P>0$
A proof of mine:
Initially, note that P is the product of dot products of vectors $\vec u_1=(1,m_1)$, $\vec u_2=(1,m_2)$, $\vec u_3=(1,m_3)$, which are director vectors of the lines BC, CA, AB: $$P=(\vec u_1.\vec u_2)(\vec u_1.\vec u_3)(\vec u_2.\vec u_3)$$
Lemma: No matter which director vectors of the sides of the triangle we choose, we are bound to find this product P with the same sign (negative, positive) or zero. After picking another director vectors, if P equals to zero, it will remain equals to zero; if P is positive, it will keep being positive; if P is negative, it will stay negative. In other words, the sign of P is invariant. Let's prove it.
Picking another director vectors $\vec u_1^{'}$,$\vec u_2^{'}$,$\vec u_3^{'}$ of lines BC, CA, AB respectively, let $$P^{'}=(\vec u_1^{'}.\vec u_2^{'})(\vec u_1^{'}.\vec u_3^{'})(\vec u_2^{'}.\vec u_3^{'})$$
As vectors $\vec u_1^{'}$ and $\vec u_1$ are vector directors of the same line (BC), they are parallel to each other and linear dependent. Therefore it exists a scalar, a real number $k_1\neq 0$, such that $\vec u_1^{'}=k_1\vec u_1$. Likewise we get $\vec u_2^{'}=k_2\vec u_2$ and $\vec u_3^{'}=k_3\vec u_3$ with $k_2\neq 0$ and $k_3\neq 0$.
Then, as $P^{'}=(\vec u_1^{'}.\vec u_2^{'})(\vec u_1^{'}.\vec u_3^{'}).(\vec u_2^{'}.\vec u_3^{'})$, $$P^{'}=((k_1\vec u_1).(k_2\vec u_2))((k_1\vec u_1).(k_3\vec u_3)).((k_2\vec u_2).((k_3\vec u_3))$$ $$P^{'}=(k_1k_2(\vec u_1.\vec u_2))(k_1k_3(\vec u_1.\vec u_3)).(k_2k_3(\vec u_2.\vec u_3))$$ $$P^{'}= (k_1k_2k_3)^2(\vec u_1.\vec u_2)(\vec u_1.\vec u_3).(\vec u_2.\vec u_3)$$ $$P^{'}= (k_1k_2k_3)^2P$$
As $k_1k_2k_3\neq 0$ and $(k_1k_2k_3)^2>0$, $P$ and $P^{'}$ are bound to be either both positive, both negative, or both null. We've thus managed to prove the invariance of the sign of the product of dot products of director vectors of the equations of the sides of a triangle, no matter which choice of them we make: $$\begin{cases} P^{'}=0 \Leftrightarrow P=0 \\ P^{'}<0 \Leftrightarrow P<0 \\ P^{'}>0 \Leftrightarrow P>0 \end{cases}$$
Having proved that, let $\vec e_1$,$\vec e_2$, $\vec e_3$ be unit vectors of the plane of the triangle ABC such that $\vec e_1$, $\vec e_2$, $\vec e_3$ have respectively the same direction and sense of vectors $\vec {BC}$, $\vec {CA}$, $\vec {AB}$, and let $$P^{'}=(\vec e_1.\vec e_2)(\vec e_1.\vec e_3)(\vec e_2.\vec e_3)$$
Then $$P^{'}=(-\cos C)(-\cos B)(-\cos A)$$ $$P^{'}=-\cos C\cos B\cos A$$
Therefore:
I. triangle ABC is right angled $\Leftrightarrow$ one of its internal angles is a right one $\Leftrightarrow$ one of the cosines of its angles is zero $\Leftrightarrow$ $P^{'}=0$ $\Leftrightarrow$ $P=0$
II. triangle ABC is acute angled $\Leftrightarrow$ all its internal angles are lesser than $\pi/2$ $\Leftrightarrow$ all cosines of its internal angles are positive $\Leftrightarrow$ the product of all these cosines is positive $\Leftrightarrow$ $P^{'}<0$ $\Leftrightarrow$ $P<0$
III. triangle ABC is obtuse angled $\Leftrightarrow$ one of its internal angles is greater than $\pi/2$ $\Leftrightarrow$ one of the cosines of its internal angles is negative $\Leftrightarrow$ the product of all these cosines is negative $\Leftrightarrow$ $P^{'}>0$ $\Leftrightarrow$ $P>0$
Arriving at these conclusions, we've finished the proof of the theorem:
Being $m_1,m_2, m_3$ the slopes of the cartesian equation of sides BC, CA, AB of a triangle ABC and $P=(1+m_1m_2)(1+m_1m_3)(1+m_2m_3)$,
I.The triangle is a right angled triangle if and only if $P=0$
II.The triangle is an acute angled triangle if and only if $P<0$
III.The triangle is an obtuse angled triangle if and only if $P>0$
Corollaries:
i) If all the slopes of the equations of the sides of a triangle have the same sign, the triangle is an obtuse angled one.
ii) If the absolute values of all the slopes of the equations of the sides of a triangle are greater than unity, the triangle is an obtuse angled one.
iii) If the absolute values of all the slopes of the equations of the sides of a triangle are lesser than unity, the triangle is an obtuse angled one.
Is This proof correct?
Is anyone acquainted with a different or alternative proof?
Does anyone know a book in which this theorem is stated and proved? I haven't found any.
