How to Prove these Areas are Equal?

This was a question from the British Mathematical Olympiad (Round 1 2007).

In the convex quadrilateral ABCD, points M,N lie on the side AB such that AM = MN =NB, and points P, Q lie on the side CD such that CP = PQ = QD. Prove that Area of AMCP = Area of MNPQ = 1/3 Area of ABCD.

I was stuck on it for some time… until I realised I had misread the question and was looking at the wrong quadrilaterals 😅. I have provided a diagram so you don’t make the same mistake.

There are no doubt several ways to prove the pink and blue quadrilaterals have equal area. But I think this is a nice method. Hopefully you agree.

Start with the blue quadrilateral AMCP, and split it into two triangles.

Now focus on the triangle ACP. If we take segment PC as its base and a perpendicular from point A to line DC as its height, we can see that it must have the same area as triangles APQ and AQD.

That’s good, because we can do the same trick for the 3 triangles CBN, CNM and CMA, which make up the rest of the quadrilateral.

We have shown that the entire quadrilateral ABCD consists of
3 × Area(ACP) + 3 × Area(CMA).

Therefore quadrilateral AMCP must contain 1/3 of the total area of quadrilateral ABCD.

Now we need to show that quadrilateral MNPQ has the same area.

Removing triangles AQD on the left and CBN on the right (which together make up 1/3 of the area), we must have 2/3 of the total area inside quadrilateral ANCQ.