Say the triangle is ABC. Let the midpoint of AB be P and the midpoint of AC be Q. Drop perpendiculars from P to BC, meeting at point R, and from Q to BC, meeting at point S. Then the rectangle PQSR has half the area of the triangle.
This one is a killer. If someone can prove it, i can get extra credit in my calc class!
"Prove that for any triangle, the inscribed rectangle with a base along one of the triangle's sides will have an area equal to half of the triangle's area."
I know that it's easy to prove visually by folding a piece of paper to look like that above, but algebraically, i don't know how.