Let $ABC$ be a triangle with $\angle B = 90^{\circ}$. Let $AD$ be the bisector of $\angle A$ with $D$ on $BC$. Suppose $AC = 6 \text{ cm}$ and the area of the $\triangle ADC$ is $10 \text{ cm}^2$. Then,the length of $BD$ in $\text{cm}$ is equal to
- A$\frac{3}{5}$
- B$\frac{3}{10}$
- C$\frac{5}{3}$
- D$\frac{10}{3}$
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