Draw AK is the altitude of triangle ABC(\(K\in BC\)),the altitude of triangle ADC from point C cut AD at I

So \(BK=CK=\dfrac{1}{2}BC\)

Applying Pitago's theorem we have:

 \(AK^2+CK^2=AC^2\Rightarrow AK^2=AC^2-CK^2=AC^2-\left(\dfrac{1}{2}BC\right)^2=6^2-3^2=27\)

\(\Rightarrow AK=3\sqrt{3}\)\(\Rightarrow S_{ADC}=\dfrac{AK.CD}{2}=\dfrac{3\sqrt{3}.4}{2}=6\sqrt{3}\)

Because BD+DK=BK \(\Rightarrow DK=BK-BD=3-2=1\)

Applying Pitago's theorem in triangle ADK we have:

 \(AD^2=AK^2+DK^2=27+1^2=28\)\(\Rightarrow AD=2\sqrt{7}\)

But \(S_{ADC}=6\sqrt{3}\)\(\Rightarrow\dfrac{AD.CH}{2}=6\sqrt{3}\Rightarrow AD.CH=12\sqrt{3}\Rightarrow CH=\dfrac{12\sqrt{3}}{2\sqrt{7}}=\dfrac{6\sqrt{21}}{7}\)