$ABCD$ is a parallelogram such that $L$ is the mid-point of $BC$. Then,$\vec{AL}$ is equal to:

  • A
    $\vec{DC} + \frac{1}{2} \vec{AD}$
  • B
    $\frac{1}{2} \vec{AD} + \vec{BC}$
  • C
    $\frac{1}{2} \vec{AD} + \vec{DL}$
  • D
    $\frac{1}{2} \vec{AD} + \vec{BL}$

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