Mathematica为什么算不出最后的结论

dlsh

从上面的程序中的结果人工变形后，可以利用向量商证明有关结论。当然，纯几何方法不难。
\(\frac{b - d}{b - c}=\frac{1}{(-1 + a o2)^2}，可变换为\frac{b - d}{b - c}=\frac{1}{a^2(1/a -o2)^2}=\frac{b^2}{(b -o2)^2}\)
即\(\frac{\overrightarrow{BD}}{\overrightarrow{BC}}=(\frac{\overrightarrow{O_1B}}{\overrightarrow{O_2B}})^2,易于推出比值等于半径平方，∠DBC=2∠O_2BO_1\)

TSC999

由于 mathematica 处理复数的能力十分有限，有时算式中不可避免的会有根式，咋办？ 我想到一个折中的方法，就是直角坐标与复数坐标相结合。
程序运行后所显示的坐标虽然是复数，但是实部与虚部是分开的。 mathematica 对于这种复数的处理能力还是不错的，有根式也不怕，下面的程序不用 1 秒钟就可运行完毕。
为什么要结合复数？因为坐标的复数表达式有许多是已知的，其表达式很简单。下面程序中的复数坐标公式是由楼主 dlsh 提供的，程序中直接引用了。
关于程序中的共轭复数，并不需要对每一个点都计算其共轭复数，只有当算式中出现共轭复数时，才有必要计算其共轭复数，否则无须计算。

1. Clear["Global`*"];
   
2. XA = (-R^2 + u^2 + 1)/(
   
3. 2 u); YA = Sqrt[-R^4 + 2 R^2 (u^2 + 1) - (u^2 - 1)^2]/(2 u); a =
   
4. XA + I YA;
   
5. XB = (-R^2 + u^2 + 1)/(
   
6. 2 u); YB = -(Sqrt[-R^4 + 2 R^2 (u^2 + 1) - (u^2 - 1)^2]/(2 u)); b =
   
7. XB + I YB;
   
8. \!\(\*OverscriptBox[\(b\), \(_\)]\) = XB - I YB;
   
9. XO1 = 0; YO1 = 0;  XO2 = u; YO2 = 0;  o1 = XO1 + I YO1; o2 =
   
10. XO2 + I YO2;
    
11. 
    
12. d = Simplify[(a - u)/(1 - a u)] ;
    
13. XD = Factor@
    
14.     Together@
    
15.      ComplexExpand@
    
16.       Re[d /. Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2] -> v] /.
    
17.    v -> Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2];
    
18. YD = Factor@
    
19.     Together@
    
20.      ComplexExpand@
    
21.       Im[d /. Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2] -> v] /.
    
22.    v -> Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2];
    
23. d = Simplify[XD] + I Simplify[YD];
    
24. c = Simplify[a - a^2 u + u];
    
25. XC = Factor@
    
26.     Together@
    
27.      ComplexExpand@
    
28.       Re[c /. Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2] -> v] /.
    
29.    v -> Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2];
    
30. YC = Factor@
    
31.     Together@
    
32.      ComplexExpand@
    
33.       Im[c /. Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2] -> v] /.
    
34.    v -> Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2];
    
35. c = Simplify[XC] + I Simplify[YC];
    
36. e = Simplify[(c - u)/(b (
    
37. \!\(\*OverscriptBox[\(b\), \(_\)]\) - u))];
    
38. XE = Factor@
    
39.     Together@
    
40.      ComplexExpand@
    
41.       Re[e /. Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2] -> v] /.
    
42.    v -> Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2];
    
43. YE = Factor@
    
44.     Together@
    
45.      ComplexExpand@
    
46.       Im[e /. Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2] -> v] /.
    
47.    v -> Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2];
    
48. e = Simplify[XE] + I Simplify[YE];
    
49. f = Simplify[b d (
    
50. \!\(\*OverscriptBox[\(b\), \(_\)]\) - u) + u];
    
51. XF = Factor@
    
52.     Together@
    
53.      ComplexExpand@
    
54.       Re[f /. Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2] -> v] /.
    
55.    v -> Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2];
    
56. YF = Factor@
    
57.     Together@
    
58.      ComplexExpand@
    
59.       Im[f /. Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2] -> v] /.
    
60.    v -> Sqrt[-R^4 + 2 (u^2 + 1) R^2 - (u^2 - 1)^2];
    
61. f = Simplify[XF] + I Simplify[YF];
    
62. m = Simplify@
    
63.    Solve[{(y - YA)/(x - XA) == (y - YD)/(x - XD), (y - YE)/(
    
64.       x - XE) == (y - YO1)/(x - XO1)}, {x, y}];  
    
65. XM = Part[Part[Part[m, 1], 1], 2];
    
66. YM = Part[Part[Part[m, 1], 2], 2];
    
67. m = XM + I YM;
    
68. n = Simplify@
    
69.    Solve[{(y - YA)/(x - XA) == (y - YC)/(x - XC), (y - YF)/(
    
70.       x - XF) == (y - YO2)/(x - XO2)}, {x, y}];
    
71. XN = Part[Part[Part[n, 1], 1], 2];
    
72. YN = Part[Part[Part[n, 1], 2], 2];
    
73. n = XN + I YN;
    
74. q = Simplify@
    
75.    Solve[{x == XA, (y - YM)/(x - XM) == (y - YN)/(x - XN)}, {x, y}];
    
76. XQ = Part[Part[Part[q, 1], 1], 2];
    
77. YQ = Part[Part[Part[q, 1], 2], 2];
    
78. q = XQ + I YQ;
    
79. Print["a = ", a]; Print["b = ", b];
    
80. Print["c = ", c]; Print["d = ", d];
    
81. Print["e = ", e]; Print["f = ", f];
    
82. Print["m = ", m]; Print["n = ", n];
    
83. Print["q = ", q];
    
84. DB = Simplify[Sqrt[(XD - XB)^2 + (YD - YB)^2]]; BC =
    
85. Simplify[Sqrt[(XC - XB)^2 + (YC - YB)^2]];
    
86. MQ = Simplify[Sqrt[(XM - XQ)^2 + (YM - YQ)^2]]; NQ =
    
87. Simplify[Sqrt[(XN - XQ)^2 + (YN - YQ)^2]];
    
88. Print["DB = ", DB]; Print["BC = ", BC];
    
89. Print["MQ = ", MQ]; Print["NQ = ", NQ];
    
90. k1 = Simplify[DB/BC, a > 0 && R > 0]; k2 =
    
91. Simplify[MQ/NQ, a > 0 && R > 0];
    
92. Print["DB/BC = ", k1]; Print["MQ/NQ = ", k2];
    
93. If[k1 == k2,
    
94. Print["由于二者之比都等于 1/\!\(\*SuperscriptBox[\(R\), \(2\)]\)，所以DB/BC = \
    
95. MQ/NQ"]]
    

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程序运行结果：

1. a = (-R^2+u^2+1)/(2 u)+(I Sqrt[-R^4+2 (u^2+1) R^2-(u^2-1)^2])/(2 u)
   
2. 
   
3. b = (-R^2+u^2+1)/(2 u)-(I Sqrt[-R^4+2 (u^2+1) R^2-(u^2-1)^2])/(2 u)
   
4. 
   
5. c = (I (R-u) Sqrt[-R^4+2 (u^2+1) R^2-(u^2-1)^2] (R+u))/(2 u)+(-R^4+2 u^2 R^2+R^2-u^4+3 u^2)/(2 u)
   
6. 
   
7. d = ((u^2-1)^2-R^2 (u^2+1))/(2 R^2 u)-(I (u-1) (u+1) Sqrt[-R^4+2 (u^2+1) R^2-(u^2-1)^2])/(2 R^2 u)
   
8. 
   
9. e = (I (R^2-u^2+u) Sqrt[-R^4+2 (u^2+1) R^2-(u^2-1)^2] (R^2-u (u+1)))/(2 R^2 u)+(-R^6+(3 u^2+1) R^4+(u^2-3 u^4) R^2+u^2 (u^2-1)^2)/(2 R^2 u)
   
10. 
    
11. f = -((I (-u^2+R u+1) Sqrt[-R^4+2 (u^2+1) R^2-(u^2-1)^2] (u^2+R u-1))/(2 R^2 u))-(u^2 R^4-(2 u^4+u^2-1) R^2+(u^2-1)^3)/(2 R^2 u)
    
12. 
    
13. m = (I Sqrt[-(R-u-1) (R-u+1) (R+u-1) (R+u+1)] (R^4-2 u^2 R^2+u^4-u^2))/(2 u (R^4-(2 u^2+1) R^2+(u^2-1)^2))+(-R^6+(3 u^2+1) R^4+(u^2-3 u^4) R^2+u^2 (u^2-1)^2)/(2 u (R^4-(2 u^2+1) R^2+(u^2-1)^2))
    
14. 
    
15. n = (I Sqrt[-(R-u-1) (R-u+1) (R+u-1) (R+u+1)] (u^4-(R^2+2) u^2+1))/(2 u (R^4-(2 u^2+1) R^2+(u^2-1)^2))+(u^2 R^4-(2 u^4+3 u^2+1) R^2+(u^2-1)^2 (u^2+1))/(2 u (R^4-(2 u^2+1) R^2+(u^2-1)^2))
    
16. 
    
17. q = (-R^2+u^2+1)/(2 u)+(I Sqrt[-(R-u-1) (R-u+1) (R+u-1) (R+u+1)] (R^6-2 u^2 R^4+u^2 (u^2-2) R^2+(u^2-1)^2))/(2 (R^2+1) u (R^4-(2 u^2+1) R^2+(u^2-1)^2))
    
18. 
    
19. DB = Sqrt[-((R^4-2 (u^2+1) R^2+(u^2-1)^2)/(R^2 u^2))]
    
20. 
    
21. BC = Sqrt[-((R^2 (R^4-2 (u^2+1) R^2+(u^2-1)^2))/u^2)]
    
22. 
    
23. MQ = Sqrt[-((R^2 (2 R^6-(5 u^2+2) R^4+(4 u^4-6 u^2-2) R^2-(u^2-2) (u^2-1)^2))/((R^2+1)^2 (R^4-(2 u^2+1) R^2+(u^2-1)^2)^2))]
    
24. 
    
25. NQ = Sqrt[-((R^6 (2 R^6-(5 u^2+2) R^4+(4 u^4-6 u^2-2) R^2-(u^2-2) (u^2-1)^2))/((R^2+1)^2 (R^4-(2 u^2+1) R^2+(u^2-1)^2)^2))]
    
26. 
    
27. DB/BC = 1/R^2
    
28. 
    
29. MQ/NQ = 1/R^2
    
30. 
    
31. 由于二者之比都等于 1/R^2，所以DB/BC = MQ/NQ

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creasson

基于有理参数表示的做法

1. 
   
2. Clear["Global`*"];
   
3. (*设点*)
   
4. points = {A->0, B->1,P->(1+I p)/(1-I u), T->(1+I q)/(1-I v)};
   
5. (*在A处的切向量*)
   
6. vectors = Factor[{v1 -> Limit[D[P/.points, u]*u^2,u->\[Infinity]],v2 -> Limit[D[T/.points, v]*v^2,v->\[Infinity]]}];
   
7. (*求C,D*)
   
8. CDEQ = (Factor[ComplexExpand[Im[#]]]//Numerator)&/@({(T-A)/v1, (P-A)/v2}/.points/.vectors);
   
9. CDsol = Solve[CDEQ == 0, {u, v}]//Flatten;
   
10. points = Union[points, {C->(T/.points/.CDsol), D->(P/.points/.CDsol)}]//Factor;
    
11. (*求E,F*)
    
12. EFEQ = (Factor[ComplexExpand[Im[#]]]//Numerator)&/@({(P-B)/(C-B), (T-B)/(D-B)}/.points);
    
13. EFsol = Solve[EFEQ == 0, {u, v}][[-1]];
    
14. points = Union[points, {E->(P/.points/.EFsol), F->(T/.points/.EFsol)}]//Factor;
    
15. (*求M,N*)
    
16. points = Union[points, {O1->(1+I p)/2, O2->(1+I q)/2}]//Factor;
    
17. points = Union[points, {M->\[Lambda]*D+(1-\[Lambda])A, N->\[Mu]*C+(1-\[Mu])A}/.points]//Factor;
    
18. MNEQ = (Factor[ComplexExpand[Im[#]]]//Numerator)&/@({(M-O1)/(E-O1), (N-O2)/(F-O2)}/.points);
    
19. MNsol = Solve[MNEQ == 0, {\[Lambda], \[Mu]}]//Flatten;
    
20. points = Factor[points/.MNsol];
    
21. (*求点Q*)
    
22. points = Union[points, {Q->\[Eta]}]//Factor;
    
23. QEQ = (Factor[ComplexExpand[Im[#]]]//Numerator)&/@({(M-Q)/(N-Q)}/.points);
    
24. Qsol = Solve[QEQ == 0, {\[Eta]}]//Factor//Flatten;
    
25. points = Factor[points/.Qsol];
    
26. (*结论*)
    
27. squarelens = Factor[ComplexExpand[Abs[#]^2]]&/@({B-D,B-C,Q-M,Q-N}/.points);
    
28. target = Factor[(squarelens[[1]]/squarelens[[2]])-(squarelens[[3]]/squarelens[[4]])];
    
29. (*显示*)
    
30. Print["points = ", points];
    
31. Print["target = ", target];
    

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TSC999

creasson 发表于 2021-10-6 23:55
基于有理参数表示的做法

如果各点的算式表达的是点的复数坐标值， 为什么与图形的实测坐标值不符合？ 程序中的 u  代表什么？ 设定了 A 在坐标原点、B 的坐标是 (1, 0) 时，是否还应该再设两个圆半径为已知（或一个圆半径及两圆心的距离），才能确定其它点的坐标？
例如将  a = 0; b = 1; p = 1.145;  q = 0.393;  代入 c = (p - q)/(p + I)  中，算出 c=0.372579 -0.325397 I， 不等于实际值  c=1.167+1.165 I。

creasson

TSC999 发表于 2021-10-7 10:52
如果各点的算式表达的是点的复数坐标值， 为什么与图形的实测坐标值不符合？ 程序中的 u  代表什么？  ...

我用的是#11的图，$p, q$为任意实数， 点 $P, T$分别表示圆$O_1,O_2$上的任意点，替换为其他字母如$X,Y$ 就好。  
半径$r_1= \frac{1}{2}\sqrt {1 + p^2} $,   $r_2 = \frac{1}{2}\sqrt {1 + q^2} $。

本图的话，若指定A=0,B=1,C=1.167+1.165I, 则由$ (p - q)/(p + I) = 1.167+1.165I$ 解出
$ p =  - 1.00172,  q = 1.33229 $

TSC999

creasson 发表于 2021-10-7 11:45
我用的是#11的图，$p, q$为任意实数， 点 $P, T$分别表示圆$O_1,O_2$上的任意点，替换为其他字母如$X,Y ...

噢，我明白了。这个设点的方法很妙啊，设了点 A、B 的坐标外，为了确定两个圆的位置以及避免运算中有根式出现，不设两个圆半径的大小，而是设 O1 的纵坐标为 p/2,  O2 的纵坐标为  q/2, 也就是设 p、q  为已知实数。其余各点的坐标都是 p、q 的函数。

经验算，用您的各点公式求出的复数坐标值全都符合 14# 页中的图。

1. Clear["Global`*"];
   
2. a = 0; b = 1; p = -2*0.518; q = 2*0.649;
   
3. c = C -> (p - q)/(p + I)
   
4. D -> -((p - q)/(q + I))
   
5. E -> ((p - I) (p - q))/((p + I) (q - I))
   
6. F -> -(((p - q) (q - I))/((p - I) (q + I)))
   
7. M -> ((p - I) (p + I) (p - q) (q - I))/(
   
8. q^2 p^2 - 3 p^2 + 8 q p - 3 q^2 + 1)
   
9. N -> -(((p - I) (p - q) (q - I) (q + I))/(
   
10.   q^2 p^2 - 3 p^2 + 8 q p - 3 q^2 + 1))
    
11. Q -> -(((p^2 + 1) (p - q)^2 (q^2 + 1))/((p^2 + q^2 + 2) (q^2 p^2 -
    
12.      3 p^2 + 8 q p - 3 q^2 + 1)))
    
13. 
    
14. C->1.16627 +1.12574 I
    
15. D->1.1284 -0.869337 I
    
16. E->0.828902 -1.15843 I
    
17. F->0.791031 +1.41483 I
    
18. M->0.387154 -0.298269 I
    
19. N->0.400147 +0.386242 I
    
20. Q->0.392816

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dlsh

主贴还是线性构造命题，用代数证明并不难，下面一段选自李涛的博士毕业论文，也是线性构造，按部就班就行了。

结论

难题

这些结论如果用纯几何方法是很难的。
一些非线性命题，看起来构造不难，反而很难证明。例如：
1  如果三角形内角平分线相等，则它是等腰三角形。
2正方形与心脏线
正△ABC中，D、E分别是AC、BC上的一点，有AD=CE，BD与AE相交于F。∠AEC和∠CBD的平分线交点G恰好位于AC上。求证：∠ABG=∠BGF。

TSC999

前面 creasson 先生程序是很正宗的机器证明程序。一时也学不会这个程序的编写方法，但是可以用直角坐标法来写程序，好处是编写容易，按相同套路一直走下去即可。

1. Clear["Global`*"];  (*令O1圆圆心的纵坐标为 p/2，O2圆圆心的纵坐标为 q/2*)
   
2. (*与以往的程序不同，点的复数坐标都采用大写字母表示。这样有其好处，比如图中有 P 这个点，所以如果 P 点的复坐标用小写字母 p \
   
3. 表示，那 O1 的纵坐标就不能用 p/2 表示了。当然点的复坐标都用大写字母表示也有坏处，那就是变量不得使用大写的 \
   
4. C、D、N，这几个大写字母是 mathematica 禁用的，因此这些点的复坐标可写成 CC、DD、NN，但用字符显示时仍可写 C、D、N \
   
5. *)
   
6. XA = 0; YA = 0; XB = 1; YB = 0; XO1 = 1/2; YO1 = p/2; XO2 = 1/2; YO2 =
   
7.   q/2;
   
8. A = XA + I YA; B = XB + I YB;  O1 = XO1 + I YO1; O2 = XO2 + I YO2;
   
9. CC = Solve[{(x - 1/2)^2 + (y - q/2)^2 == (1/2)^2 + (q/2)^2(*O2圆方程*),
   
10.     y == -1/p x(*AC方程*)}, {x, y}];
    
11. XC = Part[Part[Part[CC, 2], 1], 2];
    
12. YC = Part[Part[Part[CC, 2], 2], 2];
    
13. CC = Factor@Simplify[XC + I YC];
    
14. DD = Solve[{(x - 1/2)^2 + (y - p/2)^2 == (1/2)^2 + (p/2)^2(*O1圆方程*),
    
15.     y == -1/q x(*AD方程*)}, {x, y}];
    
16. XD = Part[Part[Part[DD, 2], 1], 2];
    
17. YD = Part[Part[Part[DD, 2], 2], 2];
    
18. DD = Factor@Simplify[XD + I YD];
    
19. Print["C = ", CC];
    
20. Print["D = ", DD];
    
21. EE = Solve[{(x - 1/2)^2 + (y - p/2)^2 == (1/2)^2 + (p/2)^2(*O1圆方程*), (
    
22.      y - YC)/(x - XC) == (y - YB)/(x - XB)(*CB方程*)}, {x, y}];
    
23. XE = Part[Part[Part[EE, 1], 1], 2];
    
24. YE = Part[Part[Part[EE, 1], 2], 2];
    
25. EE = Factor@Simplify[XE + I YE];
    
26. Print["E = ", EE];
    
27. F = Solve[{(x - 1/2)^2 + (y - q/2)^2 == (1/2)^2 + (q/2)^2(*O2圆方程*), (
    
28.      y - YD)/(x - XD) == (y - YB)/(x - XB)(*DB方程*)}, {x, y}];
    
29. XF = Part[Part[Part[F, 1], 1], 2];
    
30. YF = Part[Part[Part[F, 1], 2], 2];
    
31. F = Factor@Simplify[XF + I YF];
    
32. Print["F = ", F];
    
33. M = Solve[{(y - YA)/(x - XA) == (y - YD)/(x - XD)(*AD方程*), (y - YE)/(
    
34.      x - XE) == (y - YO1)/(x - XO1)(*EO1方程*)}, {x, y}];
    
35. XM = Part[Part[Part[M, 1], 1], 2];
    
36. YM = Part[Part[Part[M, 1], 2], 2];
    
37. M = Factor@Simplify[XM + I YM];
    
38. Print["M = ", M];
    
39. NN = Solve[{(y - YA)/(x - XA) == (y - YC)/(x - XC)(*AC方程*), (y - YF)/(
    
40.      x - XF) == (y - YO2)/(x - XO2)(*FO2方程*)}, {x, y}];
    
41. XN = Part[Part[Part[NN, 1], 1], 2];
    
42. YN = Part[Part[Part[NN, 1], 2], 2];
    
43. NN = Factor@Simplify[XN + I YN];
    
44. Print["N = ", NN];
    
45. Q = Solve[{(y - YA)/(x - XA) == (y - YB)/(x - XB)(*AB方程*), (y - YM)/(
    
46.      x - XM) == (y - YN)/(x - XN)(*MN方程*)}, {x, y}];
    
47. XQ = Part[Part[Part[Q, 1], 1], 2];
    
48. YQ = Part[Part[Part[Q, 1], 2], 2];
    
49. Q = Factor@Simplify[XQ + I YQ];
    
50. Print["Q = ", Q];
    
51. R1 = Simplify[Sqrt[(1/2)^2 + (p/2)^2]];
    
52. R2 = Simplify[Sqrt[(1/2)^2 + (q/2)^2]];
    
53. Print["R1 = ", R1];
    
54. Print["R2 = ", R2];
    
55. k = Simplify[R1^2/R2^2];
    
56. k1 = ExpandDenominator@
    
57.    Together@ComplexExpand@Simplify[Abs[(DD - B)/(B - CC)]];
    
58. k2 = ExpandDenominator@
    
59.    Together@ComplexExpand@Simplify[Abs[(M - Q)/(NN - Q)]];
    
60. Print["DB/BC = ", k1];
    
61. Print["MQ/NQ = ", k2];
    
62. Print["(R1/R2\!\(\*SuperscriptBox[\()\), \(2\)]\) = ", k];
    
63. If[k1 == k2 == k,
    
64. Print["由于 DB/BC = MQ/NQ = (R1/R2\!\(\*SuperscriptBox[\()\), \
    
65. \(2\)]\)，所以 DB/BC = MQ/NQ"]]

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程序运行结果：

1. C = (p-q)/(p+I)
   
2. D = (q-p)/(q+I)
   
3. E = ((p-I) (p-q))/((p+I) (q-I))
   
4. F = -(((p-q) (q-I))/((p-I) (q+I)))
   
5. M = ((p-I) (p+I) (p-q) (q-I))/(q^2 p^2-3 ^2+8 q p-3 q^2+1)
   
6. N = -(((p-I) (p-q) (q-I) (q+I))/(q^2 p^2-3 p^2+8 q p-3 q^2+1))
   
7. Q = -(((p^2+1) (p-q)^2 (q^2+1))/((p^2+q^2+2) (q^2 p^2-3 p^2+8 q p-3 q^2+1)))
   
8. R1 = Sqrt[p^2+1]/2
   
9. R2 = Sqrt[q^2+1]/2
   
10. DB/BC = (p^2+1)/(q^2+1)
    
11. MQ/NQ = (p^2+1)/(q^2+1)
    
12. (R1/R2)^2 = (p^2+1)/(q^2+1)
    
13. 由于 DB/BC = MQ/NQ = (R1/R2)^2，所以 DB/BC = MQ/NQ

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