Midpoint Theorem (Proof)

Proof:
\[\angle DCE = \angle ACB \;{\rm{ (common)}}\]
\[\frac{{CD}}{{CA}} = \frac{{CE}}{{CB}} = \frac{1}{2}\]
\[\Delta CDE \;{\rm{ is \;similar \;to }}\; \Delta CAB. \;{\rm{ (SAS)}}\]
Therefore,
\[\frac{{DE}}{{AB}} = \frac{1}{2}\]
And,
\[\angle CDE = \angle CAB\]
Hence,
\[DE\parallel AB\]