step 1
In the right triangle ABD
Applying the Pythagorean Theorem
![\begin{gathered} AB^2=AD^2+BD^2 \\ AB^2=3^2+BD^2 \\ AB^2=9+BD^2\text{ ----> equation 1} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tu5d87nk4nuqwjsknxvciew34px5n7j24e.png)
step 2
In the right triangle BCD
Applying the Pythagorean Theorem
![\begin{gathered} BC^2=DC^2+BD^2 \\ BC^2=12^2+BD^2 \\ BC^2=144+BD^2\text{ ----> equation 2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9zr5e74hdqt7otb1mr9umef41lsgzyuvwz.png)
step 3
In the right triangle ABC
Applying the Pythagorean Theorem
![\begin{gathered} AC^2=AB^2+BC^2 \\ 15^2=AB^2+BC^2 \\ 225=AB^2+BC^2\text{ ----> equation 3} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/smj9jn3bxezt9jfak6dwj9atvb4ya93h6z.png)
substitute equation 1 and equation 2 in equation 3
![225=(9+BD^2)+(144+BD^2)](https://img.qammunity.org/2023/formulas/mathematics/college/m3m0sz4ahjm5z69o5hfn30apz02dmev3oo.png)
Solve for BD
![\begin{gathered} \begin{equation*} 225=(9+BD^2)+(144+BD^2) \end{equation*} \\ 225-153=2BD^2 \\ BD^2=(72)/(2) \\ \\ BD^2=36 \\ BD=6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ixsjs53yzeph2ednvgxm4rq9244i3j9l0m.png)
The answer is the option C