直线交点在圆上，实在不知道程序错在哪里

dlsh

下面是用向量商的结果

1. 
   
2. Clear["Global`*"](*v=-1/a^2;*)
   
3. b = a^2 v^2;(*这句是根据AD=AE判定O1在角DAE平分线上，由Factor[k[o1,a]^2-a^2b c]可以推出*)
   
4. 
   
5. \!\(\*OverscriptBox["o", "_"]\) = o = 0;
   
6. \!\(\*OverscriptBox["c", "_"]\) = c = 1;
   
7. \!\(\*OverscriptBox["a", "_"]\) = 1/a; b = a^2 v^2;
   
8. \!\(\*OverscriptBox["b", "_"]\) = 1/b;
   
9. \!\(\*OverscriptBox["v", "_"]\) = 1/v; h = a + b + c;
   
10. \!\(\*OverscriptBox["h", "_"]\) =
    
11. \!\(\*OverscriptBox["a", "_"]\) +
    
12. \!\(\*OverscriptBox["b", "_"]\) +
    
13. \!\(\*OverscriptBox["c", "_"]\); m = (a + h)/2;
    
14. \!\(\*OverscriptBox["m", "_"]\) = (
    
15. \!\(\*OverscriptBox["a", "_"]\) +
    
16. \!\(\*OverscriptBox["h", "_"]\))/2;
    
17. o1 = (h - a v)/(1 - v);
    
18. \!\(\*OverscriptBox["o1", "_"]\) = (
    
19. \!\(\*OverscriptBox["h", "_"]\) -
    
20. \!\(\*OverscriptBox["a", "_"]\)
    
21. \!\(\*OverscriptBox["v", "_"]\))/(1 -
    
22. \!\(\*OverscriptBox["v", "_"]\));(*假设
    
23. \!\(\*OverscriptBox["O1H", "\[RightVector]"]\)/
    
24. \!\(\*OverscriptBox["O1A", "\[RightVector]"]\)=v*)
    
25. k[a_, b_] := (a - b)/(
    
26. \!\(\*OverscriptBox["a", "_"]\) -
    
27. \!\(\*OverscriptBox["b", "_"]\));
    
28. \!\(\*OverscriptBox["k", "_"]\)[a_, b_] := 1/k[a, b];(*复斜率定义*)
    
29. 
    
30. \!\(\*OverscriptBox["Jd", "_"]\)[k1_, a1_, k2_, a2_] := -((a1 - k1
    
31. \!\(\*OverscriptBox["a1", "_"]\) - (a2 - k2
    
32. \!\(\*OverscriptBox["a2", "_"]\)))/(k1 - k2));
    
33. (*复斜率等于k1,过点A1与复斜率等于k2,过点A2的直线交点*)
    
34. Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1
    
35. \!\(\*OverscriptBox["a1", "_"]\)) - k1 (a2 - k2
    
36. \!\(\*OverscriptBox["a2", "_"]\)))/(k1 - k2));
    
37. FourPoint[a_, b_, c_, d_] := ((
    
38. \!\(\*OverscriptBox["c", "_"]\) d - c
    
39. \!\(\*OverscriptBox["d", "_"]\)) (a - b) - (
    
40. \!\(\*OverscriptBox["a", "_"]\) b - a
    
41. \!\(\*OverscriptBox["b", "_"]\)) (c - d))/((a - b) (
    
42. \!\(\*OverscriptBox["c", "_"]\) -
    
43. \!\(\*OverscriptBox["d", "_"]\)) - (
    
44. \!\(\*OverscriptBox["a", "_"]\) -
    
45. \!\(\*OverscriptBox["b", "_"]\)) (c - d));(*过两点A和B、C和D的交点*)
    
46. 
    
47. \!\(\*OverscriptBox["FourPoint", "_"]\)[a_, b_, c_, d_] := -(((c
    
48. \!\(\*OverscriptBox["d", "_"]\) -
    
49. \!\(\*OverscriptBox["c", "_"]\) d) (
    
50. \!\(\*OverscriptBox["a", "_"]\) -
    
51. \!\(\*OverscriptBox["b", "_"]\)) - ( a
    
52. \!\(\*OverscriptBox["b", "_"]\) -
    
53. \!\(\*OverscriptBox["a", "_"]\) b) (
    
54. \!\(\*OverscriptBox["c", "_"]\) -
    
55. \!\(\*OverscriptBox["d", "_"]\)))/((a - b) (
    
56. \!\(\*OverscriptBox["c", "_"]\) -
    
57. \!\(\*OverscriptBox["d", "_"]\)) - (
    
58. \!\(\*OverscriptBox["a", "_"]\) -
    
59. \!\(\*OverscriptBox["b", "_"]\)) (c - d)));
    
60. Duichengdian[a_, b_, p_] := (
    
61. \!\(\*OverscriptBox["a", "_"]\) b - a
    
62. \!\(\*OverscriptBox["b", "_"]\) +
    
63. \!\(\*OverscriptBox["p", "_"]\) (a - b))/(
    
64. \!\(\*OverscriptBox["a", "_"]\) -
    
65. \!\(\*OverscriptBox["b", "_"]\));(*P关于AB的对称点*)
    
66. 
    
67. \!\(\*OverscriptBox["Duichengdian", "_"]\)[a_, b_, p_] := (a
    
68. \!\(\*OverscriptBox["b", "_"]\) -
    
69. \!\(\*OverscriptBox["a", "_"]\) b + p (
    
70. \!\(\*OverscriptBox["a", "_"]\) -
    
71. \!\(\*OverscriptBox["b", "_"]\)))/(a - b);
    
72.    (* o1=Jd[ - b c,m,-a t,a];
    
73. \!\(\*OverscriptBox["o1", "_"]\)=
    
74. \!\(\*OverscriptBox["Jd", "_"]\)[- b c,m,-a t,a];o1=Jd[a b,m,-a t,a];
    
75. \!\(\*OverscriptBox["o1", "_"]\)=
    
76. \!\(\*OverscriptBox["Jd", "_"]\)[a b,m,-a t,a];*)
    
77. d = a b (
    
78. \!\(\*OverscriptBox["a", "_"]\) -
    
79. \!\(\*OverscriptBox["o1", "_"]\)) + o1;
    
80. \!\(\*OverscriptBox["d", "_"]\) =
    
81. \!\(\*OverscriptBox["a", "_"]\)
    
82. \!\(\*OverscriptBox["b", "_"]\) (a - o1) +
    
83. \!\(\*OverscriptBox["o1", "_"]\); e = a c (
    
84. \!\(\*OverscriptBox["a", "_"]\) -
    
85. \!\(\*OverscriptBox["o1", "_"]\)) + o1;
    
86. \!\(\*OverscriptBox["e", "_"]\) =
    
87. \!\(\*OverscriptBox["a", "_"]\)
    
88. \!\(\*OverscriptBox["c", "_"]\) (a - o1) +
    
89. \!\(\*OverscriptBox["o1", "_"]\);
    
90. x = Duichengdian[o, o1, a];
    
91. \!\(\*OverscriptBox["x", "_"]\) =
    
92. \!\(\*OverscriptBox["Duichengdian", "_"]\)[o, o1, a];
    
93. q = -(k[x, d]/x);
    
94. \!\(\*OverscriptBox["q", "_"]\) = 1/q; p = -(k[x, e]/x);
    
95. \!\(\*OverscriptBox["p", "_"]\) = 1/p;
    
96. s = FourPoint[q, e, p, d];
    
97. \!\(\*OverscriptBox["s", "_"]\) =
    
98. \!\(\*OverscriptBox["FourPoint", "_"]\)[q, e, p, d];
    
99. s1 = -k[q, e] (
    
100. \!\(\*OverscriptBox["e", "_"]\) -
     
101. \!\(\*OverscriptBox["o1", "_"]\)) + o1; s2 = -k[p, d] (
     
102. \!\(\*OverscriptBox["d", "_"]\) -
     
103. \!\(\*OverscriptBox["o1", "_"]\)) + o1;
     
104. Simplify[{1, o1,
     
105. \!\(\*OverscriptBox["o1", "_"]\), , d,
     
106. \!\(\*OverscriptBox[
     
107. RowBox[{"d", "\[IndentingNewLine]"}], "_"]\), e,
     
108. \!\(\*OverscriptBox[
     
109. RowBox[{"e", "\[IndentingNewLine]"}], "_"]\), , x,
     
110. \!\(\*OverscriptBox["x", "_"]\)}]
     
111. Simplify[{2, p,
     
112. \!\(\*OverscriptBox["p", "_"]\), , q,
     
113. \!\(\*OverscriptBox[
     
114. RowBox[{"q", "\[IndentingNewLine]"}], "_"]\), , s,
     
115. \!\(\*OverscriptBox["s", "_"]\)}]
     
116. Simplify[{3, k[s, e], -((s - o1)/(
     
117. \!\(\*OverscriptBox["e", "_"]\) -
     
118. \!\(\*OverscriptBox["o1", "_"]\))), , k[s, d], k[p, d], -((s - o1)/(
     
119. \!\(\*OverscriptBox["d", "_"]\) -
     
120. \!\(\*OverscriptBox["o1", "_"]\))), -((d - o1)/(
     
121. \!\(\*OverscriptBox["s", "_"]\) -
     
122. \!\(\*OverscriptBox["o1", "_"]\))), , -((s - o1)/(
     
123. \!\(\*OverscriptBox["d", "_"]\) -
     
124. \!\(\*OverscriptBox["o1", "_"]\))) + (d - o1)/(
     
125. \!\(\*OverscriptBox["s", "_"]\) -
     
126. \!\(\*OverscriptBox["o1", "_"]\))}](*验证S在圆O1上*)
     
127. Simplify[{4, (s - o1) (
     
128. \!\(\*OverscriptBox["s", "_"]\) -
     
129. \!\(\*OverscriptBox["o1", "_"]\)), (x - o1) (
     
130. \!\(\*OverscriptBox["x", "_"]\) -
     
131. \!\(\*OverscriptBox["o1", "_"]\)), (o1 - a) (
     
132. \!\(\*OverscriptBox["o1", "_"]\) -
     
133. \!\(\*OverscriptBox["a", "_"]\)), , (e - o1) (
     
134. \!\(\*OverscriptBox["e", "_"]\) -
     
135. \!\(\*OverscriptBox["o1", "_"]\)), (d - o1) (
     
136. \!\(\*OverscriptBox["d", "_"]\) -
     
137. \!\(\*OverscriptBox["o1", "_"]\)), x
     
138. \!\(\*OverscriptBox["x", "_"]\)}]
     
139. Simplify[{5, (s - o1) (
     
140. \!\(\*OverscriptBox["s", "_"]\) -
     
141. \!\(\*OverscriptBox["o1", "_"]\)) - (x - o1) (
     
142. \!\(\*OverscriptBox["x", "_"]\) -
     
143. \!\(\*OverscriptBox["o1", "_"]\))}]
     
144. Simplify[{6, k[a, d], k[a, e], -((a - o1)/(
     
145. \!\(\*OverscriptBox["d", "_"]\) -
     
146. \!\(\*OverscriptBox["o1", "_"]\)))}]
     
147. Simplify[{7, s1, s2}]
     
148. Simplify[k[o1, a]^2 - a^2 b c]
     
149. Factor[k[o1, a]^2 - a^2 b c]
     
150. 
     
151. 
     
152. 
     

复制代码

TSC999

收到了您发给我邮箱的邮件。还没有来得及看各个附件的内容。关于向量商、共轭比的说法，我也觉得不如称为复斜率更简单、清晰，容易被人接受。您以前那篇论文，内容太少了，完全能把内容扩充十倍以上，主要是列出用于机器证明的公式，构图的方法。
我对复斜率的理解见下图。

dlsh

虽然主贴程序错误，却意外发现一条结论:
已知 △ ABC 的外接圆为O，垂心为 H ，过 A , H 两点的圆O1分别再交 AB , AC 于点 D 和 E ，且有 AD =AE． 圆O1和圆O的另一个交点为 X , XD 和 XE 分别交再圆 O于点Q和P．设M是AH中点，过M作AB垂线交∠BAC的平分线于O1’，以O1'为圆心，O1'A为半径交AB、AC于D'和E'，X'是圆O1和圆O的另一个交点。如红色圆和红色线段所示。证明： QE 和P D 的交点 S 在圆O上，X'、D'、Q和X'、E’、P共线．
以上证明包括T关于O的对称点T1的情形，如主贴所示。

TSC999

在原题中还可推出：

① A、O、S 三点共线。 ② AO 是角 PAQ 的角平分线。 ③  S 是三角形 APQ 的垂心。 ④ AP=AQ，SP=SQ，即三角形 APQ 和  SPQ  都是等腰三角形。

⑤ 以 AS 为直径的圆（圆心为 O3）交 AQ 于 D3，交 AP 于 E3，则该圆与圆 O 相内切于 A 点。D、D3、S、P 共线，E、E3、S、Q 共线。  

TSC999

补充：在图二中，证明 X2、D2、Q 三点共线； X2、E2、P 三点共线。

1. Clear["Global`*"]; (*设三角形ABC的外接圆为单位圆，C 点在正 x 轴上，如图一*)
   
2. 
   
3. \!\(\*OverscriptBox[\(o\), \(_\)]\) = o = 0;
   
4. \!\(\*OverscriptBox[\(c\), \(_\)]\) = c = 1;
   
5. 
   
6. \!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/a; b = t^2;
   
7. \!\(\*OverscriptBox[\(b\), \(_\)]\) = 1/b;
   
8. \!\(\*OverscriptBox[\(t\), \(_\)]\) = 1/t;
   
9. \!\(\*OverscriptBox[\(p\), \(_\)]\) = 1/p;
   
10. h = a + b + c(*当三角形外心在坐标原点时此式成立*);
    
11. \!\(\*OverscriptBox[\(h\), \(_\)]\) =
    
12. \!\(\*OverscriptBox[\(a\), \(_\)]\) +
    
13. \!\(\*OverscriptBox[\(b\), \(_\)]\) +
    
14. \!\(\*OverscriptBox[\(c\), \(_\)]\);
    
15. m = (a + h)/2;(* M是A、H的中点*)
    
16. \!\(\*OverscriptBox[\(m\), \(_\)]\) = (
    
17. \!\(\*OverscriptBox[\(a\), \(_\)]\) +
    
18. \!\(\*OverscriptBox[\(h\), \(_\)]\))/2;
    
19. k[a_, b_] := (a - b)/(
    
20. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
21. \!\(\*OverscriptBox[\(b\), \(_\)]\));
    
22. \!\(\*OverscriptBox[\(k\), \(_\)]\)[a_, b_] := 1/k[a, b];(*直线的复斜率*)
    
23. Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1
    
24. \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2
    
25. \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
    
26. 
    
27. \!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1
    
28. \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2
    
29. \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
    
30. (*复斜率等于k1且经过A1点的直线，与复斜率等于k2且经过A2点的直线，两直线的交点*)
    
31. FourPoint[a_, b_, c_, d_] := ((
    
32. \!\(\*OverscriptBox[\(c\), \(_\)]\) d - c
    
33. \!\(\*OverscriptBox[\(d\), \(_\)]\)) (a - b) - (
    
34. \!\(\*OverscriptBox[\(a\), \(_\)]\) b - a
    
35. \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d))/((a - b) (
    
36. \!\(\*OverscriptBox[\(c\), \(_\)]\) -
    
37. \!\(\*OverscriptBox[\(d\), \(_\)]\)) - (
    
38. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
39. \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d));
    
40. 
    
41. \!\(\*OverscriptBox[\(FourPoint\), \(_\)]\)[a_, b_, c_, d_] := -(((c
    
42. \!\(\*OverscriptBox[\(d\), \(_\)]\) -
    
43. \!\(\*OverscriptBox[\(c\), \(_\)]\) d) (
    
44. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
45. \!\(\*OverscriptBox[\(b\), \(_\)]\)) - ( a
    
46. \!\(\*OverscriptBox[\(b\), \(_\)]\) -
    
47. \!\(\*OverscriptBox[\(a\), \(_\)]\) b) (
    
48. \!\(\*OverscriptBox[\(c\), \(_\)]\) -
    
49. \!\(\*OverscriptBox[\(d\), \(_\)]\)))/((a - b) (
    
50. \!\(\*OverscriptBox[\(c\), \(_\)]\) -
    
51. \!\(\*OverscriptBox[\(d\), \(_\)]\)) - (
    
52. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
53. \!\(\*OverscriptBox[\(b\), \(_\)]\)) (c - d)));
    
54. (*过A、B点的直线与过C、D点的直线的交点*)
    
55. Duichengdian[a_, b_, p_] := (
    
56. \!\(\*OverscriptBox[\(a\), \(_\)]\) b - a
    
57. \!\(\*OverscriptBox[\(b\), \(_\)]\) +
    
58. \!\(\*OverscriptBox[\(p\), \(_\)]\) (a - b))/(
    
59. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
60. \!\(\*OverscriptBox[\(b\), \(_\)]\));(*P关于直线AB的镜像点*)
    
61. 
    
62. \!\(\*OverscriptBox[\(Duichengdian\), \(_\)]\)[a_, b_, p_] := (a
    
63. \!\(\*OverscriptBox[\(b\), \(_\)]\) -
    
64. \!\(\*OverscriptBox[\(a\), \(_\)]\) b + p (
    
65. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
66. \!\(\*OverscriptBox[\(b\), \(_\)]\)))/(a - b);
    
67. o1 = Simplify[ Jd[-k[a, h], m, k[a, t], a]];
    
68. 
    
69. \!\(\*OverscriptBox[\(o1\), \(_\)]\) = Simplify[
    
70. \!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-k[a, h], m, k[a, t], a]];
    
71. Print["o1=", o1];
    
72. w1 = Simplify@Solve[{(z - o1) (
    
73. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
    
74. \!\(\*OverscriptBox[\(o1\), \(_\)]\)) == (a - o1) (
    
75. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
76. \!\(\*OverscriptBox[\(o1\), \(_\)]\)), (z - a)/(
    
77. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
    
78. \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (z - b)/(
    
79. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
    
80. \!\(\*OverscriptBox[\(b\), \(_\)]\)), z != a}, {z,
    
81. \!\(\*OverscriptBox[\(z\), \(_\)]\)}];
    
82. d = Part[Part[Part[w1, 1], 1], 2];
    
83. 
    
84. \!\(\*OverscriptBox[\(d\), \(_\)]\) = Part[Part[Part[w1, 1], 2], 2];
    
85. Print["d = ", d];
    
86. w2 = Simplify@Solve[{(z - o1) (
    
87. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
    
88. \!\(\*OverscriptBox[\(o1\), \(_\)]\)) == (a - o1) (
    
89. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
    
90. \!\(\*OverscriptBox[\(o1\), \(_\)]\)), (z - a)/(
    
91. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
    
92. \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (z - c)/(
    
93. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
    
94. \!\(\*OverscriptBox[\(c\), \(_\)]\)), z != a}, {z,
    
95. \!\(\*OverscriptBox[\(z\), \(_\)]\)}];
    
96. e = Part[Part[Part[w2, 1], 1], 2];
    
97. 
    
98. \!\(\*OverscriptBox[\(e\), \(_\)]\) = Part[Part[Part[w2, 1], 2], 2];
    
99. Print["e=", e];
    
100. x = Simplify[Duichengdian[o, o1, a]];
     
101. \!\(\*OverscriptBox[\(x\), \(_\)]\) = Simplify[
     
102. \!\(\*OverscriptBox[\(Duichengdian\), \(_\)]\)[o, o1, a]];
     
103. Print["x=", x];
     
104. q = Simplify[-(k[x, d]/x)];
     
105. \!\(\*OverscriptBox[\(q\), \(_\)]\) = Simplify[1/q]; p =
     
106. Simplify[-(k[x, e]/x)];
     
107. \!\(\*OverscriptBox[\(p\), \(_\)]\) = Simplify[1/p];
     
108. Print["p=", p];
     
109. Print["q=", q];
     
110. s = Simplify[FourPoint[q, e, p, d]];
     
111. \!\(\*OverscriptBox[\(s\), \(_\)]\) = Simplify[
     
112. \!\(\*OverscriptBox[\(FourPoint\), \(_\)]\)[q, e, p, d]];
     
113. Print["s=", s];
     
114. Print["SE的复斜率/SD的复斜率 = ", Simplify[k[s, e]/k[s, d]]]
     
115. Print["AE的复斜率/AD的复斜率 = ", Simplify[k[a, e]/k[a, d]]]
     
116. If[Simplify[k[s, e]/k[s, d]] == Simplify[k[a, e]/k[a, d]],
     
117. Print["由于上述两个复斜率的比值相等，因此 ADSE 四点共圆。"]]
     
118. Print["OA的复斜率 = ", Simplify[k[o, a]], "     OS的复斜率 = ",
     
119.   Simplify[k[o, s]]];
     
120. If[Simplify[k[o, a]/k[o, s]] == 1,
     
121. Print["因为OA的复斜率与OS的复斜率相等，故 A、O、S 三点共线"]]
     
122. Print["AS的复斜率 = ", Simplify[k[a, s]], "     PQ的复斜率 = ",
     
123.   Simplify[k[p, q]]];
     
124. If[Simplify[k[a, s]/k[p, q]] == -1,
     
125. Print["因为AS的复斜率与PQ的复斜率是相反数，故这两条直线垂直"]]
     
126. Print["PS的复斜率 = ", Simplify[k[p, s]], "     AQ的复斜率 = ",
     
127.   Simplify[k[a, q]]];
     
128. If[Simplify[k[p, s]/k[a, q]] == -1,
     
129. Print["因为PS的复斜率与AQ的复斜率是相反数，故这两条直线垂直"]]
     
130. Print["QS的复斜率 = ", Simplify[k[q, s]], "     AP的复斜率 = ",
     
131.   Simplify[k[a, p]]];
     
132. If[Simplify[k[q, s]/k[a, p]] == -1,
     
133. Print["因为QS的复斜率与AP的复斜率是相反数，故这两条直线垂直"]]
     
134. S2[a_, b_] := (a - b) (
     
135. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
136. \!\(\*OverscriptBox[\(b\), \(_\)]\)); (* A点到B点长度的平方 *)
     
137. Print["AP = ", Simplify[S2[a, p]], "    AQ = ", Simplify[S2[a, q]]];
     
138. If[Simplify[S2[a, p]/S2[a, q]] == 1, Print["因为 AP = AQ, 故二者长度相等"]]
     
139. Print["SP = ", Simplify[S2[s, p]], "    SQ = ", Simplify[S2[s, q]]];
     
140. If[Simplify[S2[s, p]/S2[s, q]] == 1, Print["因为 SP = SQ, 故二者长度相等"]]
     
141. o2 = Simplify[ Jd[-k[a, b], m, k[a, t], a]];  
     
142. \!\(\*OverscriptBox[\(o2\), \(_\)]\) = Simplify[
     
143. \!\(\*OverscriptBox[\(Jd\), \(_\)]\)[-k[a, b], m, k[a, t], a]];
     
144. Print["o2=", o2];
     
145. w3 = Simplify@Solve[{(z - o2) (
     
146. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
147. \!\(\*OverscriptBox[\(o2\), \(_\)]\)) == (a - o2) (
     
148. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
149. \!\(\*OverscriptBox[\(o2\), \(_\)]\)), (z - a)/(
     
150. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
151. \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (z - b)/(
     
152. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
153. \!\(\*OverscriptBox[\(b\), \(_\)]\)), z != a}, {z,
     
154. \!\(\*OverscriptBox[\(z\), \(_\)]\)}];
     
155. d2 = Part[Part[Part[w3, 1], 1], 2];
     
156. 
     
157. \!\(\*OverscriptBox[\(d2\), \(_\)]\) = Part[Part[Part[w3, 1], 2], 2];
     
158. Print["d2 = ", d2];
     
159. w4 = Simplify@Solve[{(z - o2) (
     
160. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
161. \!\(\*OverscriptBox[\(o2\), \(_\)]\)) == (a - o2) (
     
162. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
163. \!\(\*OverscriptBox[\(o2\), \(_\)]\)), (z - a)/(
     
164. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
165. \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (z - c)/(
     
166. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
167. \!\(\*OverscriptBox[\(c\), \(_\)]\)), z != a}, {z,
     
168. \!\(\*OverscriptBox[\(z\), \(_\)]\)}];
     
169. e2 = Part[Part[Part[w4, 1], 1], 2];
     
170. \!\(\*OverscriptBox[\(e2\), \(_\)]\) = Part[Part[Part[w4, 1], 2], 2];
     
171. Print["e2 = ", e2];
     
172. x2 = Simplify[Duichengdian[o, o2, a]];
     
173. \!\(\*OverscriptBox[\(x2\), \(_\)]\) = Simplify[
     
174. \!\(\*OverscriptBox[\(Duichengdian\), \(_\)]\)[o, o2, a]];
     
175. Print["x2=", x2];
     
176. Print["X2D2 的复斜率 = ", Simplify[k[x2, d2]], "，    D2Q 的复斜率 = ",
     
177.   Simplify[k[d2, q]]];
     
178. If[Simplify[k[x2, d2]/ k[d2, q]] == 1,
     
179. Print["由于X2D2的复斜率等于D2Q的复斜率，因此 X2、D2、Q 三点共线。"]]
     
180. Print["X2E2 的复斜率 = ", Simplify[k[x2, e2]], "，    E2P 的复斜率 = ",
     
181.   Simplify[k[e2, p]]];
     
182. If[Simplify[k[x2, e2]/ k[e2, p]] == 1,
     
183. Print["由于X2E2的复斜率等于E2P的复斜率，因此 X2、E2、P 三点共线。"]]
     
184. o3 = (a + s)/2;
     
185. \!\(\*OverscriptBox[\(o3\), \(_\)]\) = (
     
186. \!\(\*OverscriptBox[\(a\), \(_\)]\) +
     
187. \!\(\*OverscriptBox[\(s\), \(_\)]\))/2;
     
188. w5 = Simplify@Solve[{(z - o3) (
     
189. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
190. \!\(\*OverscriptBox[\(o3\), \(_\)]\)) == (a - o3) (
     
191. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
192. \!\(\*OverscriptBox[\(o3\), \(_\)]\)), (z - a)/(
     
193. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
194. \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (z - q)/(
     
195. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
196. \!\(\*OverscriptBox[\(q\), \(_\)]\)), z != a}, {z,
     
197. \!\(\*OverscriptBox[\(z\), \(_\)]\)}];
     
198. d3 = Part[Part[Part[w5, 1], 1], 2];
     
199. 
     
200. \!\(\*OverscriptBox[\(d3\), \(_\)]\) = Part[Part[Part[w5, 1], 2], 2];
     
201. Print["d3 = ", d3];
     
202. w6 = Simplify@Solve[{(z - o3) (
     
203. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
204. \!\(\*OverscriptBox[\(o3\), \(_\)]\)) == (a - o3) (
     
205. \!\(\*OverscriptBox[\(a\), \(_\)]\) -
     
206. \!\(\*OverscriptBox[\(o3\), \(_\)]\)), (z - a)/(
     
207. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
208. \!\(\*OverscriptBox[\(a\), \(_\)]\)) == (z - p)/(
     
209. \!\(\*OverscriptBox[\(z\), \(_\)]\) -
     
210. \!\(\*OverscriptBox[\(p\), \(_\)]\)), z != a}, {z,
     
211. \!\(\*OverscriptBox[\(z\), \(_\)]\)}];
     
212. e3 = Part[Part[Part[w6, 1], 1], 2];
     
213. 
     
214. \!\(\*OverscriptBox[\(e3\), \(_\)]\) = Part[Part[Part[w6, 1], 2], 2];
     
215. Print["e3 = ", e3];
     
216. Print["DD3 的复斜率 = ", Simplify[k[d, d3]], "，    D3S 的复斜率 = ",
     
217.   Simplify[k[d3, s]], "，    SP 的复斜率 = ", Simplify[k[s, p]]];
     
218. If[Simplify[k[d, d3]/ k[d3, s]] == 1 &&
     
219.   Simplify[k[d3, s]/ k[s, p]] == 1,
     
220. Print["由于DD3、D3S、SP 的复斜率都相等，因此 D、D3、S、P 四点共线。"]]
     
221. Print["EE3 的复斜率 = ", Simplify[k[e, e3]], "，    E3S 的复斜率 = ",
     
222.   Simplify[k[e3, s]], "，    SQ 的复斜率 = ", Simplify[k[s, q]]];
     
223. If[Simplify[k[e, e3]/ k[e3, s]] == 1 &&
     
224.   Simplify[k[e3, s]/ k[s, q]] == 1,
     
225. Print["由于EE3、E3S、SQ 的复斜率都相等，因此 E、E3、S、Q 四点共线。"]]
     
226. 
     

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运行结果：

1. o1=(a (t^2-t+a+1))/(a-t)
   
2. d = (a (t^3+t^2+a+1))/(a-t)
   
3. e=(a (t^3+a t+t+1))/((a-t) t)
   
4. x=-((a t (t^2-t+a+1))/(t^2+a (t^2-t+1)))
   
5. p=-(a/t)
   
6. q=-a t
   
7. s=a (-t+1-1/t)
   
8. SE的复斜率/SD的复斜率 = 1/t^2
   
9. AE的复斜率/AD的复斜率 = 1/t^2
   
10. 由于上述两个复斜率的比值相等，因此 ADSE 四点共圆。
    
11. OA的复斜率 = a^2     OS的复斜率 = a^2
    
12. 因为OA的复斜率与OS的复斜率相等，故 A、O、S 三点共线
    
13. AS的复斜率 = a^2     PQ的复斜率 = -a^2
    
14. 因为AS的复斜率与PQ的复斜率是相反数，故这两条直线垂直
    
15. PS的复斜率 = -a^2 t     AQ的复斜率 = a^2 t
    
16. 因为PS的复斜率与AQ的复斜率是相反数，故这两条直线垂直
    
17. QS的复斜率 = -(a^2/t)     AP的复斜率 = a^2/t
    
18. 因为QS的复斜率与AP的复斜率是相反数，故这两条直线垂直
    
19. AP = t+2+1/t    AQ = t+2+1/t
    
20. 因为 AP = AQ, 故二者长度相等
    
21. SP = -((t-1)^2/t)    SQ = -((t-1)^2/t)
    
22. 因为 SP = SQ, 故二者长度相等
    
23. o2=(-a t^2+t^2+2 a t+a+1)/(2 t+2)
    
24. d2 = 1/2 (-a t^2+t^2+a+1)
    
25. e2 = (-a (t-1)^2+t^2+1)/(2 t)
    
26. x2=(t (-a t^2+t^2+2 a t+a+1))/(a t^2+t^2+2 t+a-1)
    
27. X2D2 的复斜率 = -((a t^2 (-t^2+a (t^2-2 t-1)-1))/(a t^2+t^2+2 t+a-1))，    D2Q 的复斜率 = -((a t^2 (-t^2+a (t^2-2 t-1)-1))/(a t^2+t^2+2 t+a-1))
    
28. 由于X2D2的复斜率等于D2Q的复斜率，因此 X2、D2、Q 三点共线。
    
29. X2E2 的复斜率 = (a (-a t^2+t^2+2 a t+a+1))/(a t^2+t^2+2 t+a-1)，    E2P 的复斜率 = (a (-a t^2+t^2+2 a t+a+1))/(a t^2+t^2+2 t+a-1)
    
30. 由于X2E2的复斜率等于E2P的复斜率，因此 X2、E2、P 三点共线。
    
31. d3 = -((a (t^3+t^2-t+1))/(2 t))
    
32. e3 = -((a (t^3-t^2+t+1))/(2 t^2))
    
33. DD3 的复斜率 = -a^2 t，    D3S 的复斜率 = -a^2 t，    SP 的复斜率 = -a^2 t
    
34. 由于DD3、D3S、SP 的复斜率都相等，因此 D、D3、S、P 四点共线。
    
35. EE3 的复斜率 = -(a^2/t)，    E3S 的复斜率 = -(a^2/t)，    SQ 的复斜率 = -(a^2/t)
    
36. 由于EE3、E3S、SQ 的复斜率都相等，因此 E、E3、S、Q 四点共线。

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dlsh

上面的问题属于线性构造，虽然构图复杂，计算不难，对于下面的问题如何求出u，这是四次方程。
如图，两条内角平分线相等AD=BE，求证△ABC是等腰三角形。
\(e^{\frac{i\angle BAC}{2}}=v{,}e^{\frac{i\angle CBA}{2}}=u，可求得e=\frac{\left( 1-u^2\right)v^4}{1-u^2v^4}{,}d=\frac{\left( 1-u^4\right)v^2}{1-v^2u^4}，\)\(AD=\frac{ 1-u^4}{1-u^2v^4}v,BE=\frac{ 1-v^4}{1-v^2u^4}u\)