Deriving the barycentric coordinates of a triangle's orthocenter, using the areal definition of such coordinates

Question

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.

Answer 1

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: $$ H = 3G = A+B+C.$$

Answer 2

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.