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..