1. ## Prove with vectors

Show that in any given triangle, the heights of the triangle intersect in the same (height) point..

I drew a triangle ABC and marked the intersections with heights and triangle's sides with P (on a AB side), D (BC side) and Q (AC side).
S is the intersection of the triangle's heights..

$\displaystyle \vec{AB} = \vec{a}$
$\displaystyle \vec{AC} = \vec{b}$

$\displaystyle \vec{AS} = n\vec{AD}$
$\displaystyle \vec{AD} = \vec{a} + m(\vec{b} - \vec{a})$

I have no clue what to do next..

2. Originally Posted by metlx
Show that in any given triangle, the heights of the triangle intersect in the same (height) point..

I drew a triangle ABC and marked the intersections with heights and triangle's sides with P (on a AB side), D (BC side) and Q (AC side).
S is the intersection of the triangle's heights..

$\displaystyle \vec{AB} = \vec{a}$
$\displaystyle \vec{AC} = \vec{b}$

$\displaystyle \vec{AS} = n\vec{AD}$
$\displaystyle \vec{AD} = \vec{a} + m(\vec{b} - \vec{a})$

I have no clue what to do next..
I am not sure, but think you are referring to the orthocenter. If so, this Wikipedia article shows the proof nicely. If not, could you explain in more detail the problem?

Altitude (triangle) - Wikipedia, the free encyclopedia

3. heh, that's a bit too complicated for me.. (eg. never heard for the trilinear coordinates).
I found another website with a more simple proof :P

thanks anyways