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Math  /  Geometry

QuestionComplete the proof that △EFG≅△EGF\triangle E F G \cong \triangle E G F. \begin{tabular}{|l|l|l|} \hline & Statement & Reason \\ \hline 1 & ∠F≅∠G\angle F \cong \angle G & Given \\ 2 & FG‾≅FG‾\overline{F G} \cong \overline{F G} & Reflexive Property of Congruence \\ 3 & △EFG≅△EGF\triangle E F G \cong \triangle E G F & \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. We are given that ∠F≅∠G\angle F \cong \angle G.
2. The segment FG‾\overline{FG} is congruent to itself by the Reflexive Property of Congruence.
3. We need to prove that △EFG≅△EGF\triangle EFG \cong \triangle EGF.

STEP 2

1. Identify the congruence criteria applicable to the triangles.
2. Use the given information and properties to complete the proof.

STEP 3

Identify the congruence criteria applicable to the triangles:
Since we have two angles (∠F\angle F and ∠G\angle G) that are congruent, and the side FG‾\overline{FG} is congruent to itself, we can use the Angle-Side-Angle (ASA) Congruence Postulate.

STEP 4

Use the given information and properties to complete the proof:
According to the ASA Congruence Postulate, if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.
Given: - ∠F≅∠G\angle F \cong \angle G (Given) - FG‾≅FG‾\overline{FG} \cong \overline{FG} (Reflexive Property of Congruence)
Therefore, by the ASA Congruence Postulate:
△EFG≅△EGF\triangle EFG \cong \triangle EGF
The proof is complete:
△EFG≅△EGF\boxed{\triangle EFG \cong \triangle EGF}

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