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Fild out the blanks ( blue) ,show the figureds needed  SSS Triangle Congruence Euclid's proofs of the SAS and SSS Triangle Congruence both require us to "superpose" one triangle onto another. It's never truly defined what that means, and we saw that step break in taxicab geometry. Now that we have Hilbert's Axioms for Euclidean geometry, SAS has been promoted to an axiom, and SSS can be proved using Hilbert's Axioms of Congruence.

Fild out the blanks ( blue) ,show the figureds needed  SSS Triangle Congruence Euclid's proofs of the SAS and SSS Triangle Congruence both require us to "superpose" one triangle onto another. It's never truly defined what that means, and we saw that step break in taxicab geometry. Now that we have Hilbert's Axioms for Euclidean geometry, SAS has been promoted to an axiom, and SSS can be proved using Hilbert's Axioms of Congruence.

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Fild out the blanks ( blue) ,show the figureds needed 

SSS Triangle Congruence

Euclid's proofs of the SAS and SSS Triangle Congruence both require us to "superpose" one triangle onto another. It's never truly defined what that means, and we saw that step break in taxicab geometry. Now that we have Hilbert's Axioms for Euclidean geometry, SAS has been promoted to an axiom, and SSS can be proved using Hilbert's Axioms of Congruence.

 

--- **Theorem:** If \( \triangle ABC \) and \( \triangle DEF \) are triangles such that \( AB \cong DE \), \( BC \cong EF \), and \( AC \cong DF \), then the two triangles are congruent. **Proof:** Suppose \( \triangle ABC \) and \( \triangle DEF \) are triangles such that \( AB \cong DE \), \( BC \cong EF \), and \( AC \cong DF \). It suffices to show the angles \( \angle ACB \) and \( \angle DFE \) are congruent because we can then apply Axiom blah, with included angles at \( C \) and \( F \), to conclude that the triangles \( \triangle ABC \) and \( \triangle DEF \) are congruent. First, we will make a triangle that is congruent to triangle \( \triangle DEF \) and shares side \( AB \) with triangle \( \triangle ABC \). To do this, notice that by Axiom blah, there exists a point \( P \) such that points \( C \) and \( P \) are on opposite sides of line \( \overline{AB} \) and the angle \( \angle BAP \) is congruent to the angle \( \angle EDF \). By Axiom blah, there exists a unique point \( C' \) on ray \( \overrightarrow{AP} \) such that \( AC' \cong DF \). Now, by applying Axiom blah, with included angles at \( A \) and \( D \), we can conclude triangles \( \triangle ABC \) and \( \triangle ADEF \) are congruent. Now, since \( \triangle ABC \) is congruent to \( \triangle ADEF \), we have congruent corresponding sides \( AC \cong DF \). But also \( AC \cong DF \) by hypothesis. Therefore, by Axiom blah, we can conclude \( AC \cong AC' \). Write a similar argument for why \( BC \cong BC' \). **Diagrams:** - Two triangles are drawn side by side. Triangle \( \triangle ABC \) and \( \triangle DEF \), respectively. Each has labeled points for vertices. - An additional segment \( \overline{C'P} \) is shown as a construction, illustrating the reasoning in the proof. ---
Transcribed Image Text:--- **Theorem:** If \( \triangle ABC \) and \( \triangle DEF \) are triangles such that \( AB \cong DE \), \( BC \cong EF \), and \( AC \cong DF \), then the two triangles are congruent. **Proof:** Suppose \( \triangle ABC \) and \( \triangle DEF \) are triangles such that \( AB \cong DE \), \( BC \cong EF \), and \( AC \cong DF \). It suffices to show the angles \( \angle ACB \) and \( \angle DFE \) are congruent because we can then apply Axiom blah, with included angles at \( C \) and \( F \), to conclude that the triangles \( \triangle ABC \) and \( \triangle DEF \) are congruent. First, we will make a triangle that is congruent to triangle \( \triangle DEF \) and shares side \( AB \) with triangle \( \triangle ABC \). To do this, notice that by Axiom blah, there exists a point \( P \) such that points \( C \) and \( P \) are on opposite sides of line \( \overline{AB} \) and the angle \( \angle BAP \) is congruent to the angle \( \angle EDF \). By Axiom blah, there exists a unique point \( C' \) on ray \( \overrightarrow{AP} \) such that \( AC' \cong DF \). Now, by applying Axiom blah, with included angles at \( A \) and \( D \), we can conclude triangles \( \triangle ABC \) and \( \triangle ADEF \) are congruent. Now, since \( \triangle ABC \) is congruent to \( \triangle ADEF \), we have congruent corresponding sides \( AC \cong DF \). But also \( AC \cong DF \) by hypothesis. Therefore, by Axiom blah, we can conclude \( AC \cong AC' \). Write a similar argument for why \( BC \cong BC' \). **Diagrams:** - Two triangles are drawn side by side. Triangle \( \triangle ABC \) and \( \triangle DEF \), respectively. Each has labeled points for vertices. - An additional segment \( \overline{C'P} \) is shown as a construction, illustrating the reasoning in the proof. ---
### Educational Content Transcription #### Diagram Explanation The image includes two separate line segments shown as polygons: 1. **Triangle Construction:** - A triangle \( \triangle ABC \) is constructed with points labeled \( A \), \( B \), and \( C \). - Another triangle \( \triangle DEF \) is displayed separately. 2. **Copied Triangle:** - \( \triangle ABC \) is mirrored as \( \triangle A'B'C' \) opposite \( \triangle ABC \). - Lines \( \overline{CC'} \) are drawn across the two triangles. ### Text Explanation Now that we have copied over \( \triangle DEF \) to \( \triangle ABC \), we will try to show --- \( \angle ACB \) is congruent to angle \( \angle A'C'B' \). Consider segment \( \overline{CC'} \). By construction, the points \( C \) and \( C' \) are on opposite sides of line \( \overline{AB} \), so the segment \( \overline{CC'} \) must intersect line \( \overline{AB} \) at some point, say \( G \). There are three cases to consider. #### Case 1: Points \( A \) and \( B \) are on opposite sides of point \( G \). - Diagram shows a triangular path marked with a point \( G \) between \( A \) and \( B \). Recall that we showed \( \triangle C \cong \triangle C' \). Thus, applying the Isosceles Triangle Theorem (I.5) to triangle \( \triangle ACC' \), we find the angles \( \angle ACG \) and \( \angle AC'G \) are congruent. Similarly, recall that we showed \( \triangle BC \cong \triangle BC' \). Thus, applying the Isosceles Triangle Theorem (I.5) to triangle \( \triangle BCC' \), we find the angles \( \angle BCG \) and \( \angle BC'G \) are congruent. But now, \[ \mu(\angle ACB) = \mu(\angle ACG) + \mu(\angle BCG) \] \[ = \mu(\angle AC'G) + \mu(\angle BC'G) \] \[ = \mu(\
Transcribed Image Text:### Educational Content Transcription #### Diagram Explanation The image includes two separate line segments shown as polygons: 1. **Triangle Construction:** - A triangle \( \triangle ABC \) is constructed with points labeled \( A \), \( B \), and \( C \). - Another triangle \( \triangle DEF \) is displayed separately. 2. **Copied Triangle:** - \( \triangle ABC \) is mirrored as \( \triangle A'B'C' \) opposite \( \triangle ABC \). - Lines \( \overline{CC'} \) are drawn across the two triangles. ### Text Explanation Now that we have copied over \( \triangle DEF \) to \( \triangle ABC \), we will try to show --- \( \angle ACB \) is congruent to angle \( \angle A'C'B' \). Consider segment \( \overline{CC'} \). By construction, the points \( C \) and \( C' \) are on opposite sides of line \( \overline{AB} \), so the segment \( \overline{CC'} \) must intersect line \( \overline{AB} \) at some point, say \( G \). There are three cases to consider. #### Case 1: Points \( A \) and \( B \) are on opposite sides of point \( G \). - Diagram shows a triangular path marked with a point \( G \) between \( A \) and \( B \). Recall that we showed \( \triangle C \cong \triangle C' \). Thus, applying the Isosceles Triangle Theorem (I.5) to triangle \( \triangle ACC' \), we find the angles \( \angle ACG \) and \( \angle AC'G \) are congruent. Similarly, recall that we showed \( \triangle BC \cong \triangle BC' \). Thus, applying the Isosceles Triangle Theorem (I.5) to triangle \( \triangle BCC' \), we find the angles \( \angle BCG \) and \( \angle BC'G \) are congruent. But now, \[ \mu(\angle ACB) = \mu(\angle ACG) + \mu(\angle BCG) \] \[ = \mu(\angle AC'G) + \mu(\angle BC'G) \] \[ = \mu(\
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