Since \(\widehat{ABC}=\widehat{BDC} \) and \(\hat{C}\) is the common angle, then \(\Delta ABC \sim \Delta BDC \quad (a.a)\).
It implies that: \(\dfrac{BD}{AB}=\dfrac{BC}{AC}=\dfrac{DC}{BC}\). From that we have: \(BC^2=AC.DC=144\Rightarrow BC=12\left(cm\right)\).
Then \(BD=\dfrac{AB.BC}{AC}=\dfrac{24.12}{16}=18\left(cm\right).\)
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