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

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SUMMARY

Points M and N, defined as the midpoints of opposite sides of parallelogram ABCD, successfully trisect diagonal BD, resulting in points R and S. The proof involves examining the proportional relationships between triangles BRN, BSC, DMS, and DAR, leading to the conclusion that AN is parallel to CM and that AM equals CN. Further analysis of triangles ABN and CMD, along with comparisons of triangles BRN with ARD and DSM, solidifies the trisection properties of the segments.

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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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