Each of the sides $AB$ and $CD$ of a convex quadrilateral $ABCD$ is divided into three equal parts, $|AE| = |EF| = |F B|$ , $|DP| = |P Q| = |QC|$. The diagonals of $AEPD$ and $FBCQ$ intersect at $M$ and $N$, respectively. Prove that the sum of the areas of $\vartriangle AMD$ and $\vartriangle BNC$ is equal to the sum of the areas of $\vartriangle EPM$ and $\vartriangle FNQ$.
Claim 1: $S(MDP)+S(NBF)=S(AME)+S(CNQ)$.
Proof: $S(ADP)+S(BFC)=\frac{1}{3}\cdot S(ADC)+\frac{1}{3}\cdot S(ABC)=\frac{1}{3}\cdot S(ABCD)=\frac{1}{3}\cdot S(BDC)+\frac{1}{3}\cdot S(ABD)=S(BCQ)+S(ADE)$ and this gives what we want.
Claim 2: $S(ABP)+S(CDF)=S(ABCD)$.
Proof: Since $3\cdot d(P,AB)=d(C,AB)+2\cdot d(D,AB)$ we get $3\cdot S(ABP)=S(ABC)+2\cdot S(ABD)$. Similarly we get $3\cdot S(CDF)=S(ADC)+2\cdot S(BDC)$. Sum of these $2$ equations gives that $3\cdot S(ABP)+3\cdot S(CDF)=3\cdot S(ABCD)$. So claim is proved.
Claim 3: $S(AMD)+S(BNC)=S(EPM)+S(FNQ)$.
Proof: $S(APE)+S(CFQ)=\frac{1}{3}\cdot S(ABP)+\frac{1}{3}\cdot S(CDF)=\frac{1}{3}\cdot S(ABCD)=S(APD)+S(CBF)$. Using Claim 1 gives the desired result.
So we are done!