Proof (reasons are given in brackets):
Consider [tex]\triangle ABC[/tex], [tex]\triangle EDC[/tex]:
[tex]\overline B \overline C = \overline D \overline C[/tex] (given)
[tex]\angle BCA = \angle DCE[/tex] (vertically opposite angles are equal)
[tex]\overline A \overline C = \overline E \overline C[/tex] (given)
[tex]\therefore \triangle ABC \cong \triangle EDC[/tex] (SAS)
Hope this helps :)
