Let $\overrightarrow{AB} = x$ and $\overrightarrow{AC} = y$. Observe that
\begin{align*}
\overrightarrow{AD} &= \overrightarrow{AC} + t\overrightarrow{CB} = y + t\overrightarrow{CB}, \\
\overrightarrow{AD} &= \overrightarrow{AB} + s\overrightarrow{BC} = x + s\overrightarrow{BC}.
\end{align*}
Note that $s\overrightarrow{BC}= \overrightarrow{BD}$ and $t\overrightarrow{CB} = \overrightarrow{CD}$. Also, clearly $\overrightarrow{BC} = - \overrightarrow{CB}$, so $y + t\overrightarrow{CB} = y - t\overrightarrow{BC}$. Using problem 1.1.10., we find that $t = 1- s$. Therefore, $$\frac{||\overrightar…