In quadrilateral $ABCD$,$AB=a$,$BC=b$,$AD=b-a$. If $M$ is the midpoint of $BC$ and $N$ is a point on $DM$ such that $DN=\left(\frac{4}{5}\right) DM$,then $5 AN=$

  • A
    $AC$
  • B
    $2 AC$
  • C
    $3 AC$
  • D
    $4 AC$

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