Do you think an orthocentre of an equilateral triangle only will divide its altitude in the ratio 2:1, then you are wrong! Please read this post.

A = ( x1 , y1) , B(0, 0) & C(x3, 0) O is the orthocenter and it divides perpendicular to BC which is AP in the ratio 2:1. That is AO / OP = 2 / 1 That is coordinates of O (x1, y1 /3) Slope of AC= (y1-0)/(x1 - X3) = y1 /(x1-x3) Slope of BH = (y1/3 -0) /(x1-0) = y1/(3*x1) AC perpendicular to BH, [y1/(x1-x3) ]*[(y1/(3x1) ]= -1 (y1) ^2=-3x1(x1-x3) =3x1(x3-x1) Take x1=3, x3=4 so y1= + / - 3 AB= Sqrt(10), AC = 3* Sqrt(2) = Sqrt(18) , BC = 4 AP = 3 AO = 2, OP = 1 Here we are discussing abpout only one altitude to get internally divided in the ratio 2:1. Thank you. This post I am able to write due to grace of Saibaba and SHree Gajanan Mahraj, Shegaon, Maharashtra.