$ABCD$ is a cyclic quadrilateral such that $AB$ is a diameter of the circle circumscribing it and $\angle ADC = 140^{\circ}$,then $\angle BAC$ is equal to (in $^{\circ}$)

  • A
    $80$
  • B
    $50$
  • C
    $40$
  • D
    $30$

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