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.

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.
