Average area of triangle formed ${1\over8}$ that of square Here's a question from my infamous probability textbook:

A point is taken at random in each of the two adjacent sides of a square. Show that the average area of the triangle formed by joining them is one eighth of the area of the square.

Intuitively, this makes sense since if I select the midpoints of two adjacent sides of the square, the triangle formed has ${1\over8}$ the area of the square. But I'm not sure how to go about averaging across all triangles i.e. what's the correct integral to setup. Any help would be well-appreciated.
EDIT: Following Ninad Munshi's hint in the comments, I got the expression$${{\int_0^1 \int_0^1 {{xy}\over2}\,\text{d}x\,\text{d}y}\over{1^2}},$$which indeed evaluates to ${1\over8}$ after a short calculation. But it feels like magic, can anyone explain in depth and conceptually why this integral gets us the desired probability?
EDIT 2: Thanks to heropup for his answer. However, I am asking for an explanation that specifically addresses my concerns in my previous edit, which his answer unfortunately does not.

Inside your square, draw the quadrilateral with vertices $(p,0)$, $(0,q)$, $(p,1)$, $(1,q)$. The triangles in the corners of the square have total area $1/2$. Proof: Draw the dashed lines below. Each triangle is half the area of the rectangle containing it, and those rectangles add up to area $1$.

Thus, the average area of these $4$ triangles is $1/8$. By the symmetry of the problem, therefore, the expected area of each of the triangles is $1/8$.
