Assignment 8: Altitudes and Orthocenters
Large Print Edition!
Given triangle ABC. Construct the Orthocenter
H. Let points D, E, and F be the feet of the perpendiculars from
A, B, and C
Area of triangle AHB = 1/2 x HF x AB=T(1)
Area of triangle BHC = 1/2 x HD x BC=T(2)
Area of triangle AHC = 1/2 x HE x AC=T(3)
Area of triangle ABC = area AHB + area BHC + area AHC
Area of triangle ABC = T
Using our notation T; gives us the following...
Clearly the area of ABC divided by itself is 1
With a little substitution we have:
And with a little more substitution this equals:
And with some cancellations we have:
This proves PART 1
Now we will prove PART 2:
From the picture of our triangle it is easy to see that:
So, we have proved PART 2!
But what if triangle ABC is obtuse?
The Orthocenter lies outside the triangle PART 1 and PART 2 (i.e. the equations we just proved) do not hold.
Let look at Orthocenter K which lies outside triangle ABC.
If we look closely we can see that C is an orthocenter for triangle AKC
Homework: Prove PART 1 and PART 2 for the triangle AKB