Points P, Q, R slide on the sides of parallelogram ABCD. Prove that, regardless of their position, the difference between the red areas and the blue area is constant.
Forget for a moment about points P and R and subdivisions due to their presence. Let RED be the area of two triangles ADQ and BCQ. Clearly RED = Area(ABCD)/2. Let BLUE stand for the (blue) pentagonal area of STUVW. By the Carpets Theorem applied to the trapezoid ABPR we observe that the pink area and the orange area (see that the Hint box is checked) are equal:
(1)
Area(RST) + BLUE + Area(PUV) = Area(ABW).
What we have to show (see that the Hint box is unchecked) is that
[RED - Area(RST) - Area(PUV) + Area(ABW)] - BLUE
is independent of the position of the points P, Q, R. But in view of (1)
RED - Area(RST) - Area(PUV) + Area(ABW) - BLUE = RED.