geometry

I is the incenter of ABC. We have CE = 12, AE = 9,
and AB = 15.
Find BC.

$\begin{array}{|rcll|} \hline \mathbf{ \dfrac{BC}{AB} } &=& \mathbf{ \dfrac{CE}{AE} } \\\\ \dfrac{BC}{15} &=& \dfrac{12}{9} \\\\ BC &=& \dfrac{15*12}{9} \\\\ \mathbf{BC} &=& \mathbf{20} \\ \hline \end{array}$