Metamath Proof Explorer


Theorem isosctrlem3

Description: Lemma for isosctr . Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017)

Ref Expression
Hypothesis isosctrlem3.1 F=x0,y0logyx
Assertion isosctrlem3 ABA0B0ABA=BAFBA=ABFB

Proof

Step Hyp Ref Expression
1 isosctrlem3.1 F=x0,y0logyx
2 simp1l ABA0B0ABA=BA
3 simp21 ABA0B0ABA=BA0
4 simp1r ABA0B0ABA=BB
5 2 4 subcld ABA0B0ABA=BAB
6 simp23 ABA0B0ABA=BAB
7 2 4 6 subne0d ABA0B0ABA=BAB0
8 1 angneg AA0ABAB0AFAB=AFAB
9 2 3 5 7 8 syl22anc ABA0B0ABA=BAFAB=AFAB
10 2 4 negsubdi2d ABA0B0ABA=BAB=BA
11 10 oveq2d ABA0B0ABA=BAFAB=AFBA
12 1cnd ABA0B0ABA=B1
13 ax-1ne0 10
14 13 a1i ABA0B0ABA=B10
15 4 2 3 divcld ABA0B0ABA=BBA
16 12 15 subcld ABA0B0ABA=B1BA
17 6 necomd ABA0B0ABA=BBA
18 4 2 3 17 divne1d ABA0B0ABA=BBA1
19 18 necomd ABA0B0ABA=B1BA
20 12 15 19 subne0d ABA0B0ABA=B1BA0
21 1 12 14 16 20 angvald ABA0B0ABA=B1F1BA=log1BA1
22 16 div1d ABA0B0ABA=B1BA1=1BA
23 22 fveq2d ABA0B0ABA=Blog1BA1=log1BA
24 23 fveq2d ABA0B0ABA=Blog1BA1=log1BA
25 4 2 3 absdivd ABA0B0ABA=BBA=BA
26 simp3 ABA0B0ABA=BA=B
27 26 eqcomd ABA0B0ABA=BB=A
28 27 oveq1d ABA0B0ABA=BBA=AA
29 2 abscld ABA0B0ABA=BA
30 29 recnd ABA0B0ABA=BA
31 2 3 absne0d ABA0B0ABA=BA0
32 30 31 dividd ABA0B0ABA=BAA=1
33 25 28 32 3eqtrd ABA0B0ABA=BBA=1
34 19 neneqd ABA0B0ABA=B¬1=BA
35 isosctrlem2 BABA=1¬1=BAlog1BA=logBA1BA
36 15 33 34 35 syl3anc ABA0B0ABA=Blog1BA=logBA1BA
37 15 negcld ABA0B0ABA=BBA
38 simp22 ABA0B0ABA=BB0
39 4 2 38 3 divne0d ABA0B0ABA=BBA0
40 15 39 negne0d ABA0B0ABA=BBA0
41 1 16 20 37 40 angvald ABA0B0ABA=B1BAFBA=logBA1BA
42 36 41 eqtr4d ABA0B0ABA=Blog1BA=1BAFBA
43 21 24 42 3eqtrd ABA0B0ABA=B1F1BA=1BAFBA
44 2 mulridd ABA0B0ABA=BA1=A
45 2 12 15 subdid ABA0B0ABA=BA1BA=A1ABA
46 4 2 3 divcan2d ABA0B0ABA=BABA=B
47 44 46 oveq12d ABA0B0ABA=BA1ABA=AB
48 45 47 eqtrd ABA0B0ABA=BA1BA=AB
49 44 48 oveq12d ABA0B0ABA=BA1FA1BA=AFAB
50 1 angcan 1101BA1BA0AA0A1FA1BA=1F1BA
51 12 14 16 20 2 3 50 syl222anc ABA0B0ABA=BA1FA1BA=1F1BA
52 49 51 eqtr3d ABA0B0ABA=BAFAB=1F1BA
53 2 15 mulneg2d ABA0B0ABA=BABA=ABA
54 46 negeqd ABA0B0ABA=BABA=B
55 53 54 eqtrd ABA0B0ABA=BABA=B
56 48 55 oveq12d ABA0B0ABA=BA1BAFABA=ABFB
57 1 angcan 1BA1BA0BABA0AA0A1BAFABA=1BAFBA
58 16 20 37 40 2 3 57 syl222anc ABA0B0ABA=BA1BAFABA=1BAFBA
59 56 58 eqtr3d ABA0B0ABA=BABFB=1BAFBA
60 43 52 59 3eqtr4d ABA0B0ABA=BAFAB=ABFB
61 9 11 60 3eqtr3d ABA0B0ABA=BAFBA=ABFB