In this investigation, I will be taking a look at the orthocenter of a triangle. I initially began my investigation by trying to prove that
(AG/AD) + (BH/BE) + (CI/CF) = 4
I spent far too much time trying to prove this with no progress. I took it to Dr. Mo Hendon and Dr. Ed Azoff for guidance and both had no clue what to do. So I gave up the ghost and went for something simpler.
I am now going to prove that the altitudes of a triangle have a concurrent point, which is known as the orthocenter.
I will start with the following arbitrary triangle.
Now I will proof the altitudes of the triangle are concurrent using linear algebra. I might as well get some use out of it since I took the class three times. Note: the * means dot product.
From my construction of the altitudes, I know that
(i) (O-A) * (B-C) = 0
(ii) => (O*B) – (O*C) – (A*B) + (A*C) = 0
(iii) (O-B) * (C-A) = 0
(iv) => (O*C) – (O*A) – (B*C) + (B*A) = 0
(v) (O-C) * (B-A) = 0
(vi) => (O*B) – (O*A) – (C*B) + (C*A) = 0
If I subtract (ii) from (iv) then I get (vi), which proves that the altitudes are concurrent.