By the geometric mean theorem we know that:
[tex]CD^2=AD\times DB[/tex]
from the pythagorean theorem we also know that:
[tex]CB^2=CD^2+DB^2[/tex]
Now, if we plug the value of CD squared from the first equation into the second one we have:
[tex]\begin{gathered} CB^2=AD\times DB+DB^2 \\ CB^2=DB(AD+DB) \\ \text{but } \\ AD+DB=AB \\ \text{then} \\ CB^2=AB\times DB \\ CB=\sqrt[]{AB\times DB} \end{gathered}[/tex]
Therefore:
[tex]CB=\sqrt[]{AB\times DB}[/tex]
