step 1
In the right triangle ABD
Applying the Pythagorean Theorem
[tex]\begin{gathered} AB^2=AD^2+BD^2 \\ AB^2=3^2+BD^2 \\ AB^2=9+BD^2\text{ ----> equation 1} \end{gathered}[/tex]
step 2
In the right triangle BCD
Applying the Pythagorean Theorem
[tex]\begin{gathered} BC^2=DC^2+BD^2 \\ BC^2=12^2+BD^2 \\ BC^2=144+BD^2\text{ ----> equation 2} \end{gathered}[/tex]
step 3
In the right triangle ABC
Applying the Pythagorean Theorem
[tex]\begin{gathered} AC^2=AB^2+BC^2 \\ 15^2=AB^2+BC^2 \\ 225=AB^2+BC^2\text{ ----> equation 3} \end{gathered}[/tex]
substitute equation 1 and equation 2 in equation 3
[tex]225=(9+BD^2)+(144+BD^2)[/tex]
Solve for BD
[tex]\begin{gathered} \begin{equation*} 225=(9+BD^2)+(144+BD^2) \end{equation*} \\ 225-153=2BD^2 \\ BD^2=\frac{72}{2} \\ \\ BD^2=36 \\ BD=6 \end{gathered}[/tex]
The answer is the option C
