[tex]\text{Given that }\Delta ABC\cong\Delta DEF[/tex]
Let us take by SSS these triangles are congruent.
So these two triangles have equal corresponding angles.
[tex]\angle ABC=\angle DEF[/tex]
Consider the triangle GEC.
[tex]\angle ACB\text{ and }\angle ECG\text{ are vertically opposite angles.}[/tex]
Recall that vertically opposite are always equal.
[tex]\angle ACB\text{=}\angle ECG\text{ }[/tex]
[tex]\angle DEF\text{ and }\angle CEG\text{ are vertically opposite angles.}[/tex]
[tex]\angle DEF\text{ =}\angle CEG[/tex]
[tex]\text{ Replace that }\angle ABC=\angle DEF\text{ .}[/tex]
[tex]\angle ABC\text{ =}\angle CEG[/tex]
Hence triangles ABC and GEC have corresponding two angles that are equal.
If two corresponding angles are equal then third must be equal.
For example
By using the triangle sum property on triangle ABC, we get
[tex]\angle ABC+\angle ACB+\angle BAC=180^o[/tex]
[tex]\angle BAC=180^o-\angle ABC-\angle ACB\text{ take it as equation (1).}[/tex]
Similarly to triangle GEC.
[tex]\angle CEG+\angle CGE+\angle ECG=180^o[/tex]
[tex]\angle CGE=180^o-\angle CEG-\angle\text{ECG}[/tex][tex]\text{ Replace }\angle CEG=\angle ABC\text{ and }\angle ECG=\angle ACB,\text{ we get}[/tex]
[tex]\angle CEG=180^o-\angle ABC-\angle ACB\text{ take it as equation (2).}[/tex]
Comparing equations (1) and (2), we get that
[tex]\angle BAC=\angle CEG[/tex]
Therefore triangles ABC and GEC are similar since their corresponding angles are congruent.
