Triangle question

[Jaspworld]

In triangle AOB, H is the point of intersection of the altitude from A and B. Let a(vector)=OA(vector), b(vector)=OB(vector), and h(vector)=OH(vector).

(Note: * means the dot product)
a.) Prove that a*(b-h)=0
b.) Prove that b*(a-h)=0
c.) Using the results of parts a and b, prove that h*(b-a)=0. Explain the geometrical significance of this result.

Just get me started.

 

[soroban]

Hello, Jaspworld!

In triangle AOB, H is the point of intersection of the altitudes from A and B.
Let \(\displaystyle a\,=\,\vec{OA},\;b\,=\,\vec{OB},\;h\,=\,\vec{OH}\)

                  O
                  *
                 /|\
                / | \
               /  |  \
              /   |   \
             /    |    \
            /     |     \
           *      |      *
          /   *   |   *   \
         /        *        \
        /     *   H   *     \
       /  *               *  \
    A * - - - - - - - - - - - * B

a ) Prove that: \(\displaystyle \:a\cdot(b\,-\,h)\:=\:0\)

We have: \(\displaystyle \:b\,-\,h\:=\:\vec{OB}\,-\vec{OH} \:=\:\vec{OB}\,+\,\vec{HO}\:=\:\vec{HB}\)

We are given that: \(\displaystyle \vec{OA}\,\perp\,\vec{HB}.\;\) Hence: \(\displaystyle \vec{OA}\,\cdot\,\vec{HB}\,=\,0\)

. . Therefore: \(\displaystyle \:a\cdot(b\,-\,h)\:=\:\vec{OA}\,\cdot\,\vec{HB} \:=\:0\)

b) Prove that: \(\displaystyle \: b\cdot(a\,-\,h)\:=\:0\)

The proof is similar to part (a).

c) Using the results of parts a and b, prove that: \(\displaystyle \:h\cdot(b\,-\,a)\:=\:0\)
Explain the geometrical significance of this result.

From (a), we have: \(\displaystyle \:a\cdot(b\,-\,h)\:=\:0\;\;\Rightarrow\;\;a\cdot h \:=\:a\cdot b\)

From (b), we have: \(\displaystyle \:b\cdot(a\,-\,h)\:-\:0\;\;\Rightarrow\;\;b\cdot h \:=\:a\cdot b\)

Hence, we have: \(\displaystyle \:b\cdot h \:=\:a\cdot h\;\;\Rightarrow\;\;b\cdot h \,-\,a\cdot h\:=\:0\)

. . Therefore: \(\displaystyle \:h\cdot(b\,-\,a)\:=\:0\)


Since \(\displaystyle b\,-\,a\:=\:\vec{OB}\,-\,\vec{OA}\:=\:\vec{OB}\,+\,\vec{AO}\:=\:\vec{AB}\),

. . we have shown that: \(\displaystyle h\,\perp\,\vec{AB}\)

That is, the line from vertex \(\displaystyle O\) through the intersection \(\displaystyle H\)
. . of the two altitudes is also an altitude.

We have shown that the three altitudes of a triangle are concurrent.
. . (The common intersection is the orthocenter.)

 

[Jaspworld]

Thanks

Thanks a lot. I'm just wondering if you know a website or a good book that teaches geometry (for example: a book with questions like this one and complete detailed solutions).
For some time, I had trouble seeing that OB+HO=HB!