In the convex quadrilateral ABCD, points M and N lie on AB such that AM =MN=NB.Points P and Q lie on side CD such that CP=PQ=QD.How can we prove that area of AMCP is 1/3 of ABCD?
Draw diagonal AC.
Show that area(AMC) = 1/3 area(ABC)
hint: think of AM and AB as the bases of the two triangles.
Do the same with CPA and CDA.