Given that ABC is an acute triangle, construct the orthocenter H. Let Points D, E, and F be the feet of the perpendiculars from A, B, and C respectively.
We should consider what happens if ABC is an obtuse
triangle, and see if the above relationships still hold.
First, we would construct the orthocenter H of triangle ABC, and let the points D, E, and F be the feet of the perpendiculars from A, B, and C respectively.
We know that
area (ABC) = area (HAB) + area (HAC) + area (HBC)
Now, divide both sides of the above equation by A, to get:
After simplifying the fractions we get:
To find the validity of the second case we know that:
Rearrangement of terms gives
When we have an obtuse triangle the orthocenter will be outside the triangle. In the figure below the orthocenter of triangle ABC is constructed. Then feet of the perpendiculars from vertices A, B, and C are points D, E, and F respectively.
In this case
area (BCH) = area (ABC) + area (AHB) + area (CAH)
After simplifying fractions we get:
As we can see this is different from equation (1) which we proved earlier.
To prove the second relationship, we know that:
Do you see any similarities between the relationships for the acute and obtuse triangles?