Intercept Theorem (Proof)

Proof:
\[\angle DCE = \angle ACB \;{\rm{ (common)}}\]
\[\angle CDE = \angle CAB\;{\rm{ (alt}}{\rm{. }}\;\angle {\rm{s)}}\]
\[\Delta CDE \;{\rm{ is \;similar \;to }}\; \Delta CAB. \;{\rm{ (AAA)}}\]
Therefore,
$$\begin{align}
\frac{{CD}}{{CA}} & = \frac{{CE}}{{CB}} \\
\\
\frac{{CD}}{{CD + AD}} & = \frac{{CE}}{{CE + BE}} \\
\\
\frac{{CD + AD}}{{CD}} & = \frac{{CE + BE}}{{CE}} \\
\\
\frac{{CD}}{{CD}} + \frac{{AD}}{{CD}} & = \frac{{CE}}{{CE}} + \frac{{BE}}{{CE}} \\
\\
1 + \frac{{AD}}{{CD}} & = 1 + \frac{{BE}}{{CE}} \\
\\
\frac{{AD}}{{CD}} & = \frac{{BE}}{{CE}} \\
\end{align}$$