Vì BD là pg nên \(\dfrac{AB}{BC}=\dfrac{AD}{DC}\Rightarrow AB.DC=AD.BC\Leftrightarrow AD.BC=80cm\)
\(AD+DC=BC\Leftrightarrow BC-AD=DC=10\)
\(\Leftrightarrow\left\{{}\begin{matrix}AD.BC=80\\BC-AD=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AD=\dfrac{80}{BC}\\BC-\dfrac{80}{BC}=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}BC=5+\sqrt{105}\\BC=5-\sqrt{105}\left(KTM\right)\end{matrix}\right.\)