Deriving the barycentric coordinates of a triangle's orthocenter, using the areal definition of such coordinates Wikipedia's "Altitude (triangle)" entry  describes the barycentric coordinates of $\triangle ABC$'s orthocenter as $$(\tan A : \tan B : \tan C)$$

How would you prove this using solely the areal definition of barycentric coordinates?

Thank you.
 A: i am going to give you a string of equalities. you need to be familliar with $\sin t = \frac{opp}{hyp}$ and $\cos t = \frac{adj}{hyp}.$  i am going to take the diameter of the circumcircle to be $2$ so that $$a = BC = \sin A, b = \sin B. c = \sin C. $$ let the orthocenter be $H, L$ the foot of perp from $A$ to $BC.$  from the right triangle $ABL,$ verify that $$BL = \sin C\cos B, BC = \cos B, HL = \cos B \cos C.$$  and twice the area of the triangle $BCH = \sin A \cos B\cos C.$
therefore we have $$area\, BCH:area\, ACH: area\, ABH = \cos B \cos C\sin A: \cos A \cos C\sin B:\cos A \cos B\sin C$$ now, divide every thing by $\cos A \cos B \cos C$ should give you what you are after.
A: 
Barycentric Coordinates
In the triangle above,
$$
\begin{align}
CE&=\tan(A)\,AE\tag{1a}\\[6pt]
DE&=\cot(B)\,AE\tag{1b}\\
\frac{DE}{CE}&=\cot(A)\cot(B)\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\triangle AEC$ has right angle at $E$
$\text{(1b)}$: $\triangle DEA\cong\triangle BGA$ and $\triangle DEA$ has a right angle at $E$
$\text{(1c)}$: divide $\text{(1b)}$ by $\text{(1a)}$
The ratio of the altitudes of $\triangle ADB$ and $\triangle ACB$ is the ratio of their areas. Therefore, the area definition of the barycentric coordinates says that the $C$-barycentric coordinate for $D$ is $\cot(A)\cot(B)$. Thus, by similarity,
$$
\begin{align}
D
&=A\,\overbrace{\cot(B)\cot(C)}^\text{$A$-coordinate}+B\,\overbrace{\cot(C)\cot(A)}^\text{$B$-coordinate}+C\,\overbrace{\cot(A)\cot(B)}^\text{$C$-coordinate}\tag{2a}\\[6pt]
&=\frac{A\,\tan(A)+B\,\tan(B)+C\,\tan(C)}{\tan(A)\tan(B)\tan(C)}\tag{2b}\\
&=\frac{A\,\tan(A)+B\,\tan(B)+C\,\tan(C)}{\tan(A)+\tan(B)+\tan(C)}\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: extend the coordinates by similarity
$\text{(2b)}$: multiply by $\frac{\tan(A)\tan(B)\tan(C)}{\tan(A)\tan(B)\tan(C)}$
$\text{(2c)}$: if $A+B+C=\pi$, then $\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$
We could stop here because $\text{(2c)}$ gives the proportions $\tan(A):\tan(B):\tan(C)$, but we can compute those tangents given the sides.

Computing the Tangents
In the triangle above, the Pythagorean Theorem says
$$
\begin{align}
\overbrace{a^2-BE^2}^{CE^2}&=\overbrace{b^2-AE^2}^{CE^2}\tag{3a}\\
AE^2-BE^2&=b^2-a^2\tag{3b}
\end{align}
$$
Explanation:
$\text{(3a)}$: Pythagorean Theorem
$\text{(3b)}$: add $AE^2-a^2$
Furthermore, 
$$
\begin{align}
AE
&=\frac12((AE-BE)+(AE+BE))\tag{4a}\\
&=\frac12\left(\frac{AE^2-BE^2}{AE+BE}+(AE+BE)\right)\tag{4b}\\
&=\frac12\left(\frac{b^2-a^2}c+c\right)\tag{4c}\\
&=\frac{b^2-a^2+c^2}{2c}\tag{4d}
\end{align}
$$
Explanation:
$\text{(4a)}$: algebra
$\text{(4b)}$: algebra
$\text{(4c)}$: apply $\text{(3b)}$ and $c=AE+BE$
$\text{(4d)}$: algebra
$$
\begin{align}
CE
&=\sqrt{b^2-AE^2}\tag{5a}\\
&=\frac{\sqrt{4b^2c^2-\left(b^2-a^2+c^2\right)^2}}{2c}\tag{5b}\\
&=\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{2c}\tag{5c}
\end{align}
$$
Explanation:
$\text{(5a)}$: Pythagorean Theorem
$\text{(5b)}$: apply $\text{(4d)}$
$\text{(5c)}$: factor
Combining $\text{(4d)}$, $\text{(5c)}$, and $\text{(1a)}$, we get $\tan(A)$, and by similarity, $\tan(B)$ and $\tan(C)$:
$$
\begin{align}
\tan(A)&=\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{b^2+c^2-a^2}\tag{6a}\\
\tan(B)&=\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{c^2+a^2-b^2}\tag{6b}\\
\tan(C)&=\frac{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}{a^2+b^2-c^2}\tag{6c}
\end{align}
$$

Computing $\boldsymbol{D}$
Applying $\text{(2c)}$ and $(6)$, we get
$$
D=\frac{\frac{A}{b^2+c^2-a^2}+\frac{B}{c^2+a^2-b^2}+\frac{C}{a^2+b^2-c^2}}{\frac1{b^2+c^2-a^2}+\frac1{c^2+a^2-b^2}+\frac1{a^2+b^2-c^2}}\tag7
$$
A: We have that the orthocenter $H$ is the isogonal conjugate of the circumcenter $O$, hence the trilinear coordinates of $H$ and $O$ are given by:
$$ O:[\cos A,\cos B,\cos C],\qquad H:[\sec A,\sec B,\sec C]$$
and the conversion between trilinear coordinates and barycentric coordinates is straightforward.
As an alternative, we may notice that $O,H$ and the centroid $G$ are collinear and fulfill $\frac{HO}{GO}=3$ by Euler's theorem; that gives that if we put the origin in $O$ we simply have:
$$ H = 3G = A+B+C.$$
