Question

gh ⊥ eh, gh ∥ ef, and ef ⊥ fg. complete the proof that △efg ≅ △ghe.

1 gh ⊥ eh
2 gh ∥ ef
3 ef ⊥ fg
4 ∠f ≅ ∠h
5 ∠feg ≅ ∠egh
6 eg ≅ eg
7 △efg ≅ △ghe

1 given
2 given
3 given
4
5
6 reflexive property of congruence
7

Explanation:

Step1: Determine the reason for $\angle FEG\cong\angle EGH$

Since \(GH\parallel EF\), by the Alternate - Interior Angles Theorem, \(\angle FEG\cong\angle EGH\).

Step2: Determine the reason for \(\triangle EFG\cong\triangle GHE\)

We have \(EF\perp FG\), \(GH\perp EH\), so \(\angle EFG = \angle GHE=90^{\circ}\). We know \(GH\parallel EF\) (given), so \(\angle FEG=\angle EGH\) (Alternate - Interior Angles Theorem). Also, \(EG = EG\) (Reflexive Property of Congruence). By the AAS (Angle - Angle - Side) Congruence Criterion, \(\triangle EFG\cong\triangle GHE\).

Answer:

The reason for \(\angle FEG\cong\angle EGH\) is "Alternate - Interior Angles Theorem" and the reason for \(\triangle EFG\cong\triangle GHE\) is "AAS (Angle - Angle - Side) Congruence Criterion".