In the essay Lockhart's Lament (page 4), the author describes a proof for the standard area of triangle $(bh)/2$ by enclosing the triangle in a rectangle and chopping the rectangle into two (perhaps unequal parts) using its altitude (which is the perpendicular from the top vertex to the base). Of the four triangles so formed, we see that they are partitioned into pairs of congruent triangles of equal area; hence the area of the triangle is precisely half of the rectangle. We assume that the altitude is lying inside the triangle here.
Now suppose the altitude lies outside the triangle. Can a similar argument still be made?
I am not looking for any rigorous proof but only a hint.
Yes - the rectangle is split into two large right triangles, and each large right triangle is split into an obtuse triangle and a small right triangle.