Proof of Triangle Congruence: SAS Postulate

Posted on Jan 21, 2025 in Mathematics

Theorem 8: Triangle Congruence via SAS Postulate

Statements

1. In \( \triangle ABC \leftrightarrow \triangle GET \)

• i) \( BC = WE \)
• ii) \( \angle B = \angle GET \)
• iii) \( BA = GE \)

\( \therefore \triangle ABC \cong \triangle GEF \) \( \therefore AC = GF \) & \( \angle A = \angle GET \) But \( DF = A \) \( \therefore GF = DF \) \( \therefore \) In \( \triangle DEG \), \( m \angle 1 = m \angle \) Similarly, in \( \triangle GFD \), \( m \angle 2 = m \angle 4 \) \( \therefore m \angle 1 + m \angle 2 = m \angle + m \angle 4 \) \( \therefore m \angle D = m \angle G \) But \( m \angle G = m \angle A \) \( \therefore m \angle A = m \angle D \) \( \therefore \) In \( \triangle ABC \leftrightarrow \triangle DEF \)

• i) \( AB = DE \)
• ii) \( \angle A = \angle D \)
• iii) \( AC = DF \)

\( \therefore \triangle ABC \cong \triangle DEF \)

Reasons

1. • i) Given
   • ii) Construction
   • iii) Construction
2. SAS Postulate
3. By the congruence of triangles
4. Given
5. Transitive property
6. Opposite sides congruent (Theorem 6)
7. \( DF = GF \) (Theorem 6)
8. Addition property of equation
9. \( m \angle 1 + m \angle 2 = m \angle D \)
10. Proved in (3) above
11. Transitive property
12. • i) Given
    • ii) Proved above
    • iii) Given
13. SAS Postulate