Metamath Proof Explorer


Theorem sigaradd

Description: Subtracting (double) area of A D C from A B C yields the (double) area of D B C . (Contributed by Saveliy Skresanov, 23-Sep-2017)

Ref Expression
Hypotheses sharhght.sigar G = x , y x y
sharhght.a φ A B C
sharhght.b φ D A D G B D = 0
Assertion sigaradd φ B C G A C D C G A C = B C G D C

Proof

Step Hyp Ref Expression
1 sharhght.sigar G = x , y x y
2 sharhght.a φ A B C
3 sharhght.b φ D A D G B D = 0
4 2 simp1d φ A
5 2 simp3d φ C
6 3 simpld φ D
7 4 5 6 nnncan1d φ A - C - A D = D C
8 7 oveq2d φ B D G A - C - A D = B D G D C
9 2 simp2d φ B
10 9 6 subcld φ B D
11 4 5 subcld φ A C
12 4 6 subcld φ A D
13 1 sigarms B D A C A D B D G A - C - A D = B D G A C B D G A D
14 10 11 12 13 syl3anc φ B D G A - C - A D = B D G A C B D G A D
15 8 14 eqtr3d φ B D G D C = B D G A C B D G A D
16 1 sigarac A D B D A D G B D = B D G A D
17 12 10 16 syl2anc φ A D G B D = B D G A D
18 3 simprd φ A D G B D = 0
19 17 18 eqtr3d φ B D G A D = 0
20 19 negeqd φ B D G A D = 0
21 10 12 jca φ B D A D
22 1 21 sigarimcd φ B D G A D
23 22 negnegd φ B D G A D = B D G A D
24 neg0 0 = 0
25 24 a1i φ 0 = 0
26 20 23 25 3eqtr3d φ B D G A D = 0
27 26 oveq2d φ B D G A C B D G A D = B D G A C 0
28 10 11 jca φ B D A C
29 1 28 sigarimcd φ B D G A C
30 29 subid1d φ B D G A C 0 = B D G A C
31 15 27 30 3eqtrd φ B D G D C = B D G A C
32 9 6 5 nnncan2d φ B - C - D C = B D
33 32 oveq1d φ B - C - D C G A C = B D G A C
34 9 5 subcld φ B C
35 6 5 subcld φ D C
36 1 sigarmf B C A C D C B - C - D C G A C = B C G A C D C G A C
37 34 11 35 36 syl3anc φ B - C - D C G A C = B C G A C D C G A C
38 31 33 37 3eqtr2rd φ B C G A C D C G A C = B D G D C
39 5 6 subcld φ C D
40 1red φ 1
41 40 renegcld φ 1
42 1 sigarls B D C D 1 B D G C D -1 = B D G C D -1
43 10 39 41 42 syl3anc φ B D G C D -1 = B D G C D -1
44 39 mulm1d φ -1 C D = C D
45 1cnd φ 1
46 45 negcld φ 1
47 46 39 mulcomd φ -1 C D = C D -1
48 5 6 negsubdi2d φ C D = D C
49 44 47 48 3eqtr3d φ C D -1 = D C
50 49 oveq2d φ B D G C D -1 = B D G D C
51 10 39 jca φ B D C D
52 1 51 sigarimcd φ B D G C D
53 52 mulm1d φ -1 B D G C D = B D G C D
54 52 46 mulcomd φ B D G C D -1 = -1 B D G C D
55 1 sigarac C D B D C D G B D = B D G C D
56 39 10 55 syl2anc φ C D G B D = B D G C D
57 53 54 56 3eqtr4d φ B D G C D -1 = C D G B D
58 43 50 57 3eqtr3d φ B D G D C = C D G B D
59 1 sigarperm C B D C D G B D = B C G D C
60 5 9 6 59 syl3anc φ C D G B D = B C G D C
61 38 58 60 3eqtrd φ B C G A C D C G A C = B C G D C