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Question
- given: \\( \overline { e g } \\) is the perpendicular bisector of \\( \overline { f h } \\) prove \\( \triangle f g e \cong \triangle h g e \\) statements justifications 1. 1. given 2. \\( \overline { f g } \cong \overline { h g } \\) 2. definition of perpendicular bisector 3. \\( \overline { e g } \cong \overline { e g } \\) 3. 4. 4. definition of perpendicular bisector 5. \\( \triangle f g e \cong \triangle h g e \\) 5.
Step1: Identify the given information
We are given that \( \overline{EG} \) is the perpendicular bisector of \( \overline{FH} \).
Step2: Analyze statement 3
For statement 3, \( \overline{EG}\cong\overline{EG} \) by the reflexive property of congruence (a segment is congruent to itself).
Step3: Analyze statement 4
Since \( \overline{EG} \) is the perpendicular bisector of \( \overline{FH} \), \( \angle EGF=\angle EGH = 90^{\circ}\) (by the definition of perpendicular bisector).
Step4: Prove the triangles congruent
We have \( \overline{FG}\cong\overline{HG}\) (from statement 2, definition of perpendicular bisector), \( \overline{EG}\cong\overline{EG}\) (reflexive property), and \( \angle EGF=\angle EGH\) (from statement 4, definition of perpendicular bisector). So, by the Side - Angle - Side (SAS) congruence criterion, \( \triangle FGE\cong\triangle HGE\).
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- \( \overline{EG} \) is the perpendicular bisector of \( \overline{FH} \)
- \( \overline{FG}\cong\overline{HG} \)
- Reflexive property of congruence
- \( \angle EGF=\angle EGH = 90^{\circ}\)
- SAS (Side - Angle - Side) congruence criterion