You also seem to be given that [tex]m\angle ADC=50^\circ[/tex], according to the diagram.
In triangle ABC, we have
[tex]\cos\angle CAB=\dfrac39=\dfrac13\implies m\angle CAB=m\angle CAD=\cos^{-1}\dfrac13[/tex]
By the law of sines,
[tex]\dfrac{\sin\angle ADC}{AC}=\dfrac{\sin\angle CAD}{CD}[/tex]
[tex]\implies CD=\dfrac{(9\,\mathrm{cm})\sin\left(\cos^{-1}\frac13\right)}{\sin50^\circ}\approx11.1\,\mathrm{cm}[/tex]
