Orthocenter of a Triangle:A different approach

In general, when we are given a triangle and we attempt to find its orthocenter, our general technique is to determine the equations of the straight lines from the co-ordinates of the triangle, and use slope and normal concepts to solve for the point.

Here is a different and simple approach of finding it...which becomes really handy sometimes...


Suppose, we have the following triangle ABC, with A=(x1,y1), B=(x2,y2) and C=(x3,y3).see the figure]. We produce three perpendicular lines from A,B,C on the respective sides, which meet on G. So, G is the Orthocenter of the triangle by definition.

Now, we make a further construction, i.e. we draw a line parallel to BC through A. Similarly we draw two more such lines through B(parallel to CA) and through C(parallel to AB). Since A,B,C are non-collinear points, these lines will intersect in three different points, suppose they are D,E,F.

So, by this new construction, what we get is a bigger triangle DEF, where A is the midpoint of DF, B is the midpoint of DE and C is the midpoint of EF.

So, now by definition, the three perpendiculars are basically the perpendicular bisectors of the sides of the triangle DEF. By definition, G is the Circumcenter of triangle DEF.

Knowing the co-ordinates of the points D,E,F will be easy once co-ordinates of A,B,C are known. Say D=(x4,y4),E=(x5,y5) and F=(x6,y6).

Thus, (x4+x6)/2 = x1, (y4+y6)/2 = y1
(x4+x5)/2 = x2, (y4+y5)/2 = y2
(x5+x6)/2 = x3, (y5+y6)/2 = y3

From here we can solve for (x4,y4),(x5,y5) and (x6,y6) easily. Once the co-ordinates of D,E,F is known, we can consider G=(h,k) and use the distance formula to calculate the circumcenter co-ordinates.

This is an alternative, and sometimes easier(based on the expressions for D,E,F) method of calculating the co-ordinates of the Orthocenter of a given triangle.

P/S- I avoided the expressions of D,E,F or G in terms of A,B,C co-ordinates as they are really clumsy. They become easier when a numerical value is supplied.