In a triangle, in which ratio orthocenter divides the altitude?

AO/OP = cos A / (cos B cos C) Using 1) A = 180 degree - (B + C) 2) cos [180 -(B + C)] = -cos(B + C) and cos (B + C) = cosBCosC-sinBsinC In triangle ABC if AP is perpendicular to BC such that B-P-C i.e. P is the foot of perpendicular from vertex A on side BC and O is orthocenter then AO/OP = tanB tan C - 1 From this we understand that in a triangle the ratio in which the orthocenter divides the altitude does not depend on the lengths of the sides but depends on the angles of the triangle. The vertex from which altitude is drawn i.e. the angle A 1) will be acute if the ratio is positive 2) will be 90 degrees if the ratio is zero 3) will be an obtuse angle if the ratio is negative. Conversely if the ratio is 1) positive the angle A will be an acute angle 2) zero angle A will be 90 degrees 3) negative the angle A will be an obtuse angle.