Triangle inequality is an exterior angle of a right triangle and therefore larger than , meaning the other base angle of the isosceles triangle also is greater than and their sum exceeds in violation of the triangle postulate). This theorem establishing inequalities is sharpened by Pythagoras' theorem to the equality that the square of the length of the hypotenuse equals the sum of the squares of the other two sides. Consider a triangle whose sides are in an arithmetic progression and let the sides be , , . Then the triangle inequality requires that The first of these quadratic inequalities requires to
Pythagorean theorem and the remaining square. Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. The details follow. Let "A", "B", "C" be the vertices of a right triangle, with a right angle at "A". Drop a perpendicular from "A" to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs. For the formal proof, we require