Metamath Proof Explorer


Theorem hypcgr

Description: If the catheti of two right-angled triangles are congruent, so is their hypothenuse. Theorem 10.12 of Schwabhauser p. 91. (Contributed by Thierry Arnoux, 16-Dec-2019)

Ref Expression
Hypotheses hypcgr.p ⊢ P = Base G
hypcgr.m ⊢ - ˙ = dist ⁡ G
hypcgr.i ⊢ I = Itv ⁡ G
hypcgr.g ⊢ φ → G ∈ 𝒢 Tarski
hypcgr.h ⊢ φ → G Dim 𝒢 ≥ 2
hypcgr.a ⊢ φ → A ∈ P
hypcgr.b ⊢ φ → B ∈ P
hypcgr.c ⊢ φ → C ∈ P
hypcgr.d ⊢ φ → D ∈ P
hypcgr.e ⊢ φ → E ∈ P
hypcgr.f ⊢ φ → F ∈ P
hypcgr.1 ⊢ φ → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G
hypcgr.2 ⊢ φ → ⟨“ DEF ”⟩ ∈ ∟ 𝒢 ⁡ G
hypcgr.3 ⊢ φ → A - ˙ B = D - ˙ E
hypcgr.4 ⊢ φ → B - ˙ C = E - ˙ F
Assertion hypcgr ⊢ φ → A - ˙ C = D - ˙ F

Proof

Step Hyp Ref Expression
1 hypcgr.p ⊢ P = Base G
2 hypcgr.m ⊢ - ˙ = dist ⁡ G
3 hypcgr.i ⊢ I = Itv ⁡ G
4 hypcgr.g ⊢ φ → G ∈ 𝒢 Tarski
5 hypcgr.h ⊢ φ → G Dim 𝒢 ≥ 2
6 hypcgr.a ⊢ φ → A ∈ P
7 hypcgr.b ⊢ φ → B ∈ P
8 hypcgr.c ⊢ φ → C ∈ P
9 hypcgr.d ⊢ φ → D ∈ P
10 hypcgr.e ⊢ φ → E ∈ P
11 hypcgr.f ⊢ φ → F ∈ P
12 hypcgr.1 ⊢ φ → ⟨“ ABC ”⟩ ∈ ∟ 𝒢 ⁡ G
13 hypcgr.2 ⊢ φ → ⟨“ DEF ”⟩ ∈ ∟ 𝒢 ⁡ G
14 hypcgr.3 ⊢ φ → A - ˙ B = D - ˙ E
15 hypcgr.4 ⊢ φ → B - ˙ C = E - ˙ F
16 eqid ⊢ Line 𝒢 ⁡ G = Line 𝒢 ⁡ G
17 eqid ⊢ pInv 𝒢 ⁡ G = pInv 𝒢 ⁡ G
18 1 2 3 4 5 7 10 midcl ⊢ φ → B mid 𝒢 ⁡ G E ∈ P
19 eqid ⊢ pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E = pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E
20 1 2 3 16 17 4 18 19 9 mircl ⊢ φ → pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ D ∈ P
21 1 2 3 16 17 4 18 19 10 mircl ⊢ φ → pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ E ∈ P
22 1 2 3 16 17 4 18 19 11 mircl ⊢ φ → pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ F ∈ P
23 1 2 3 16 17 4 9 10 11 13 19 18 mirrag ⊢ φ → ⟨“ pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ D pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ E pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ F ”⟩ ∈ ∟ 𝒢 ⁡ G
24 1 2 3 16 17 4 18 19 9 10 miriso ⊢ φ → pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ D - ˙ pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ E = D - ˙ E
25 14 24 eqtr4d ⊢ φ → A - ˙ B = pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ D - ˙ pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ E
26 1 2 3 16 17 4 18 19 10 11 miriso ⊢ φ → pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ E - ˙ pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ F = E - ˙ F
27 15 26 eqtr4d ⊢ φ → B - ˙ C = pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ E - ˙ pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ F
28 1 2 3 4 5 10 7 midcom ⊢ φ → E mid 𝒢 ⁡ G B = B mid 𝒢 ⁡ G E
29 1 2 3 4 5 10 7 17 18 ismidb ⊢ φ → B = pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ E ↔ E mid 𝒢 ⁡ G B = B mid 𝒢 ⁡ G E
30 28 29 mpbird ⊢ φ → B = pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ E
31 eqid ⊢ lInv 𝒢 ⁡ G ⁡ C mid 𝒢 ⁡ G pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ F Line 𝒢 ⁡ G B = lInv 𝒢 ⁡ G ⁡ C mid 𝒢 ⁡ G pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ F Line 𝒢 ⁡ G B
32 1 2 3 4 5 6 7 8 20 21 22 12 23 25 27 30 31 hypcgrlem2 ⊢ φ → A - ˙ C = pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ D - ˙ pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ F
33 1 2 3 16 17 4 18 19 9 11 miriso ⊢ φ → pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ D - ˙ pInv 𝒢 ⁡ G ⁡ B mid 𝒢 ⁡ G E ⁡ F = D - ˙ F
34 32 33 eqtrd ⊢ φ → A - ˙ C = D - ˙ F