Given a triangle (diagram at bottom of post).
Treating the triangular points as unit points $e_1, e_2, e_3$, and $T(e_3 - e_2)$ for example as the length of the side opposite $e_1$, I am reading an 1890 representation of Grassmann's Ausdehnungslehre (1862), in which the orthocenter $p'$ is presented as $\frac{l \cdot e_1 + m \cdot e_2 + n \cdot e_3}{l + m + n}$, where $l = \frac{T(e_3 - e_2)}{cos(\angle\ at\ e_1)}$, $m = \frac{T(e_1 - e_3)}{cos(\angle\ at\ e_2)}$, $n = \frac{T(e_2 - e_1)}{cos(\angle\ at\ e_3)}$.
In this formulation $\frac{m \cdot e_2 + n \cdot e_3}{m + n} = p_1$, $\frac{n \cdot e_3 + l \cdot e_1}{n + l} = p_2$, and $\frac{l \cdot e_1 + m \cdot e_2}{l + m} = p_3$.
Because this is an older presentation, and I am not familiar with the newer developments involving Grassmann's works, that is exterior algebra, wedge operator, Grassmann numbers, and so on, I recognize that this presentation may seem somewhat obscure but if there is anyone that can answer using the terminology of the late 1800's rather than the modern interpretation, that would be appreciated.
It is presented following a simpler but similar example for the median intersection, aka center of gravity or centroid, where the foot of each median is just the midpoint of the side, e.g. $\frac{e_2 + e_3}{2}$.
In the simpler example, the following reasoning, with $p$ the centroid, is $p = x \cdot e_1 + y \cdot p_1 = x' \cdot e_2 + y' \cdot p_2 = x \cdot e_1 + \frac{y}{2}(e_2 + e_3) = x' \cdot e_2 + \frac{y'}{2}(e_3 + e_1)$, $(x - \frac{1}{2}y')e_1 + (\frac{1}{2}y - x')e_2 + \frac{1}{2}(y - y')e_3 = 0$.
Since $m_1 \cdot p_1 + m_2 \cdot p_2 + m_3 \cdot p_3 = 0$ with $m_1 + m_2 + m_3 \neq 0$ for weights $m_1, m_2, m_3$ is a condition that the three points $p_1, p_2, p_3$ are collinear but they are not collinear, because they form a triangle, $m_1 + m_2 + m_3 = 0$, e.g. $(x - \frac{1}{2}y') = (\frac{1}{2}y - x') = \frac{1}{2}(y - y') = 0$.
But, since $p$ is a unit point $x + y = 1 = x' + y'$.
From these equations we find $x = x' = \frac{1}{3}$, $y = y' = \frac{2}{3}$,
whence $p = \frac{1}{3}e_1 + \frac{2}{3}p_1 = \frac{1}{3}(e_1 + e_2 + e_3)$ which shows that $p$ is the mean point of $e_1, e_2, e_3$ and trisects the distance from $e_1$ to $p_1$.
Although I think I understand this centroid example, I do not understand the similar construction for the orthocenter.
The solution of equations proceeds exactly the same way as the centroid example, once the expression for the orthocenter $p'$ is stated analogous to the way the expression for the centroid $p$ was stated in the previous example.
Where I get lost is the expressions that are given for $p_1, p_2, p_3$ in the orthocenter example.
$l, m, n$ are verbally described as the ratios of the sides of the triangle to the cosines of the opposite angles.
