In $$\triangle ABC, \angle C = 90^\circ$$ and $$D$$ is a point on $$CB$$ such that $$AD$$ is the bisector of $$\angle A$$. If $$AC = 5 cm$$ and $$BC = 12cm$$, then what is the length of $$AD$$?

From the above question, we draw the diagram is given below 

From the above question 

then $$\cos (\frac{A}{2}) = \frac{5}{AD}$$  [$$cos (\frac{A}{2}) = \sqrt{\frac{1 + cosA}{2}}$$]
$$\frac{5}{AD} = \sqrt {\frac{1 + cosA}{2}}$$

we know,
$$cosA = \frac{5}{13}$$
$$\frac{5}{AD} = \sqrt{\frac{1 + \dfrac{5}{13}}{2}}$$
Solving this we get,
$$\Rightarrow AD = \frac{5\sqrt{13}}{3}$$

$$\Rightarrow AD= 6.009$$

then Option (C) Ans