Can Points M and N Trisect a Parallelogram's Diagonal and Sides?

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Points M and N, as midpoints of opposite sides in parallelogram ABCD, are crucial for proving that R and S trisect diagonal BD. The discussion emphasizes examining the proportions between triangles BRN, BSC, DMS, and DAR to establish that AN is parallel to CM and that AM equals CN. Additionally, comparing triangles ABN and CMD, along with triangles BRN and ARD, supports the trisection claims. The geometric relationships and properties of parallelograms are key to the proof. The thread seeks assistance in completing these geometric proofs efficiently.
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i need to finish this before i move on tomorrow. any help? thanks

Points M and N are the midpoints of opposite sides of paralellogram ABCD. Prove that:
a) R and S trisect diagonal BD
b) R is a points of trisection of AN and S is a point of trisection of MC.

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Try examining the proportions between the triangles BRN, BSC, DMS,DAR
 
Prove AN||CM and AM = CN. Now go with triangle ABN and CMD. Then compare triangles as Pseudo said like BRN and ARD and then BRN and DSM
 
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