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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?

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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.

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  • $\begingroup$ Thanks for the answer.I am sorry i did not reply before.Was busy with fermat's lil theorem. $\endgroup$ – rah4927 Jun 8 '13 at 16:08

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