Question

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$\angle geh \cong \angle feh$ and $\angle f \cong \angle g$. complete the proof that $\overline{eg} \cong \overline{ef}$.
1 $\angle geh \cong \angle feh$ given
2 $\angle f \cong \angle g$ given
3 $\overline{eh} \cong \overline{eh}$ reflexive property of congruence
4 $\triangle efh \cong \triangle egh$
5 $\overline{eg} \cong \overline{ef}$ cpctc

Explanation:

Step1: Recall Congruence Theorems

We have two angles (\(\angle FEH\cong\angle GEH\), \(\angle F\cong\angle G\)) and a common side (\(\overline{EH}\cong\overline{EH}\)). The AAS (Angle - Angle - Side) congruence theorem states that if two angles and a non - included side of one triangle are congruent to two angles and the corresponding non - included side of another triangle, then the two triangles are congruent.

Step2: Apply the Theorem

In \(\triangle EFH\) and \(\triangle EGH\), we have \(\angle FEH\cong\angle GEH\) (given), \(\angle F\cong\angle G\) (given), and \(\overline{EH}\cong\overline{EH}\) (reflexive property). So, by AAS congruence, \(\triangle EFH\cong\triangle EGH\).

Answer:

AAS (Angle - Angle - Side) Congruence Theorem.