Eritrea's Theorem

Solution 1:

This is a different way to obtain the key equality $2a\alpha = c^2-b^2$.

We will use notion of power of a point with respect to a circle.

On the one hand, power of $G$ with respect to a circle centered at $I$ and radius $\frac b2$ equals $GI^2-\left(\frac b2\right)^2=\frac{c^2-b^2}{4}$.

On the other hand, this circle passes through $C$ and $D$ so the power of $G$ is $GD \cdot GC = \alpha \cdot \frac a2$ (assuming segments are oriented).

Thus $\frac{c^2-b^2}{4}=\alpha \cdot \frac a2$, so $2a \alpha = c^2-b^2$.

Solution 2:

Express α, β, γ in terms of a, b, c

According to the Encyclopedia of Triangle Centers, the orthocenter $X(4)$ has barycentric coordinates $[\tan A:\tan B:\tan C]$. From the cosine law you have $\cos C=\frac{a^2+b^2-c^2}{2ab}$ and likewise for the other angles. So you get

\begin{align*} \tan C&=\frac{\sin C}{\cos C}=\frac{\sqrt{1-\cos C^2}}{\cos C} =\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}{a^2+b^2-c^2} \\&=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{a^2+b^2-c^2} \\&=\frac{4V}{a^2+b^2-c^2} \end{align*}

where $V$ denotes the area of the triangle, as obtained from Heron's formula. By canceling the $4V$ term and multiplying with all the denominators (which is allowed for homogeneous coordinates), you might write the barycentric coordinates of $X(4)$ also as

$$\begin{bmatrix} (a^2-b^2+c^2)(a^2+b^2-c^2)\\ (b^2-c^2+a^2)(b^2+c^2-a^2)\\ (c^2-a^2+b^2)(c^2+a^2-b^2) \end{bmatrix}$$

You get the footpoints of the heights by setting one of these coordinates to zero. So for example $D$ has barycentric coordinates

$$\begin{bmatrix} 0\\ (b^2-c^2+a^2)(b^2+c^2-a^2)\\ (c^2-a^2+b^2)(c^2+a^2-b^2) \end{bmatrix}\sim\begin{bmatrix} 0\\ b^2-c^2+a^2\\ c^2+a^2-b^2 \end{bmatrix}$$

which means you can write its Euclidean coordinates as

\begin{align*} D&=\frac{b^2-c^2+a^2}{(b^2-c^2+a^2)+(c^2+a^2-b^2)}B +\frac{c^2+a^2-b^2}{(b^2-c^2+a^2)+(c^2+a^2-b^2)}C \\&=\frac{a^2+b^2-c^2}{2a^2}B + \frac{a^2-b^2+c^2}{2a^2}C \\&=B + \frac{a^2-b^2+c^2}{2a^2}(C-B) \\&=B + \left(\frac12 + \frac{-b^2+c^2}{2a^2}\right)(C-B) \\&=G + \frac{-b^2+c^2}{2a^2}(C-B) \end{align*}

The distance between $D$ and $G$ is

$$\alpha=\lVert D-G\rVert = \left\lvert\frac{-b^2+c^2}{2a^2}\right\rvert\cdot\lVert C-B\rVert = \frac{\lvert b^2-c^2\rvert}{2a}\\ 2a\alpha=\lvert b^2-c^2\rvert$$

Orientation and order

Up to here, the above is an alternative to the shorter deduction timon92 posted in his answer. The discussion which follows below applies no matter how one obtains that formula for $2a\alpha$.

Using this formula, the equation of the theorem (multiplied by $2$ to simplify things) would be

$$\lvert b^2-c^2\rvert + \lvert c^2-a^2\rvert = \lvert a^2-b^2\rvert$$

This is not always the case. But if you use signed distances, e.g. always measured in counter-clockwise direction, you can omit the absolute values. Then write the formula as “sum equals zero” as columbus8myhw suggests in his comment, and you obtain

$$(b^2-c^2) + (c^2-a^2) + (a^2-b^2) = 0$$

which is obviously true.

If you prefer unsigned distances, when does the equation with those hold? It holds if and only if the difference inside the absolute value function has equal sign for both terms on the left hand side of the equation but opposite sign on the right. So you have two cases to consider:

\begin{gather*} b^2-c^2\ge0,\quad c^2-a^2\ge0,\quad a^2-b^2\le0 \quad\implies\quad b\ge c\ge a \\ b^2-c^2\le0,\quad c^2-a^2\le0,\quad a^2-b^2\ge0 \quad\implies\quad a\ge c\ge b \end{gather*}

That's what the “middle side” g.kov quoted in his comment refers to: the $c$ on the right hand side of the equation must be the side of median length.