In $\Delta ABC$,$\overline{AM}$ and $\overline{CN}$ are altitudes. If $AB = 12$,$BC = 15$ and $AM = 9.6$,then $CN = \ldots$

  • A
    $12$
  • B
    $6.4$
  • C
    $7.2$
  • D
    $6$

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = 4AD^{2} - 3AB^{2}$.

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