Metamath Proof Explorer


Theorem trgcgr

Description: Triangle congruence. (Contributed by Thierry Arnoux, 27-Apr-2019)

Ref Expression
Hypotheses trgcgrg.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
trgcgrg.m ⊢ − = ( dist ‘ 𝐺 )
trgcgrg.r ⊢ ∼ = ( cgrG ‘ 𝐺 )
trgcgrg.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
trgcgrg.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
trgcgrg.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
trgcgrg.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
trgcgrg.d ⊢ ( 𝜑 → 𝐷 ∈ 𝑃 )
trgcgrg.e ⊢ ( 𝜑 → 𝐸 ∈ 𝑃 )
trgcgrg.f ⊢ ( 𝜑 → 𝐹 ∈ 𝑃 )
trgcgr.1 ⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) = ( 𝐷 − 𝐸 ) )
trgcgr.2 ⊢ ( 𝜑 → ( 𝐵 − 𝐶 ) = ( 𝐸 − 𝐹 ) )
trgcgr.3 ⊢ ( 𝜑 → ( 𝐶 − 𝐴 ) = ( 𝐹 − 𝐷 ) )
Assertion trgcgr ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 ”⟩ ∼ ⟨“ 𝐷 𝐸 𝐹 ”⟩ )

Proof

Step Hyp Ref Expression
1 trgcgrg.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
2 trgcgrg.m ⊢ − = ( dist ‘ 𝐺 )
3 trgcgrg.r ⊢ ∼ = ( cgrG ‘ 𝐺 )
4 trgcgrg.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
5 trgcgrg.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
6 trgcgrg.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
7 trgcgrg.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
8 trgcgrg.d ⊢ ( 𝜑 → 𝐷 ∈ 𝑃 )
9 trgcgrg.e ⊢ ( 𝜑 → 𝐸 ∈ 𝑃 )
10 trgcgrg.f ⊢ ( 𝜑 → 𝐹 ∈ 𝑃 )
11 trgcgr.1 ⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) = ( 𝐷 − 𝐸 ) )
12 trgcgr.2 ⊢ ( 𝜑 → ( 𝐵 − 𝐶 ) = ( 𝐸 − 𝐹 ) )
13 trgcgr.3 ⊢ ( 𝜑 → ( 𝐶 − 𝐴 ) = ( 𝐹 − 𝐷 ) )
14 1 2 3 4 5 6 7 8 9 10 trgcgrg ⊢ ( 𝜑 → ( ⟨“ 𝐴 𝐵 𝐶 ”⟩ ∼ ⟨“ 𝐷 𝐸 𝐹 ”⟩ ↔ ( ( 𝐴 − 𝐵 ) = ( 𝐷 − 𝐸 ) ∧ ( 𝐵 − 𝐶 ) = ( 𝐸 − 𝐹 ) ∧ ( 𝐶 − 𝐴 ) = ( 𝐹 − 𝐷 ) ) ) )
15 11 12 13 14 mpbir3and ⊢ ( 𝜑 → ⟨“ 𝐴 𝐵 𝐶 ”⟩ ∼ ⟨“ 𝐷 𝐸 𝐹 ”⟩ )