Here is a construction involving perpendiculars dropped from each vertice down meet the opposite side at 90 degrees. We can see that D always appears to exist. This is important because D is the intersection of all three perpendiculars. We are going to try to prove that this intersection always exists.

Theorem: If perpendiculars dropped from each vertice in a triangle ABC intersect at a point D, then this intersection always exists.
Proof: (by contradiction) Assume that D does exist. Show there exists a situation where this is not the case.
So, Perp(A) ll Perp(B) ll Perp(C) (Perp(x) = perpendicular dropped from x)
If this were the case, then 90 degrees would exist at each angle.
Not only would this mean the angle measures exceed 180, but the points would then be colinear.
So, when triangle ABC has perpendiculars dropped from each vertice to the opposite segment, D will always exist.