ΔABC 中，AC⊥CB，D,E 在 AC,AB 上，AD=CD=BE，F 是 DE 中点，求证：∠AED=∠CBF

denglongshan

直角三角形中，AD = DC = BE，F是DE中点，求证；\[Angle]AED = 角DBF。

来自纯几何吧

ccmmjj

好题，坐等几何证明。

天山草

先来个复解析法证明：

denglongshan

1. 
   
2. 
   
3. Clear["Global`*"]
   
4. 
   
5. \!\(\*OverscriptBox["a", "_"]\) = a = 0;
   
6. \!\(\*OverscriptBox["b", "_"]\) = 1/b;(*假设A在原点，AC与实轴重合，AB=1*)
   
7. 
   
8. \!\(\*OverscriptBox["c", "_"]\) = c = (b +
   
9. \!\(\*OverscriptBox["b", "_"]\))/2;
   
10. \!\(\*OverscriptBox["d", "_"]\) = d = c/2; e = (1 - d) b;
    
11. \!\(\*OverscriptBox["e", "_"]\) = (1 - d)
    
12. \!\(\*OverscriptBox["b", "_"]\); f = (d + e)/2;
    
13. \!\(\*OverscriptBox["f", "_"]\) = (
    
14. \!\(\*OverscriptBox["d", "_"]\) +
    
15. \!\(\*OverscriptBox["e", "_"]\))/2;
    
16. KAB[a_, b_] := (a - b)/(
    
17. \!\(\*OverscriptBox["a", "_"]\) -
    
18. \!\(\*OverscriptBox["b", "_"]\));
    
19. \!\(\*OverscriptBox["KAB", "_"]\)[a_, b_] := 1/KAB[a, b];(*复斜率定义*)
    
20. Angle2[a_, b_, c_] := KAB[a, b]/KAB[b, c];(*e^(2iB) 等于复斜率相除*)
    
21. Tan1[a_, b_, c_] := (KAB[a, b] - KAB[b, c])/(KAB[b, c] + KAB[a, b]) I;
    
22. 
    
23. \!\(\*OverscriptBox["Jd", "_"]\)[k1_, a1_, k2_, a2_] := -((a1 - k1
    
24. \!\(\*OverscriptBox["a1", "_"]\) - (a2 - k2
    
25. \!\(\*OverscriptBox["a2", "_"]\)))/(
    
26.   k1 - k2));(*复斜率等于k1,过点A1与复斜率等于k2,过点A2的直线交点*)
    
27. Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1
    
28. \!\(\*OverscriptBox["a1", "_"]\)) - k1 (a2 - k2
    
29. \!\(\*OverscriptBox["a2", "_"]\)))/(k1 - k2));
    
30. 
    
31. Simplify[{1, , d,
    
32. \!\(\*OverscriptBox["d", "_"]\), , e,
    
33. \!\(\*OverscriptBox["e", "_"]\), , f,
    
34. \!\(\*OverscriptBox["f", "_"]\)}]
    
35. Simplify[{2, KAB[d, e], KAB[a, e], , KAB[f, b]}]
    
36. Simplify[{3, Angle2[d, e, a], Angle2[c, b, f],
    
37.   Angle2[d, e, a] == Angle2[c, b, f]}](*平方单位验证角度相等*)
    
38. Simplify[{30, Tan1[d, e, a], Tan1[c, b, f],
    
39.   Tan1[d, e, a] == Tan1[c, b, f]}](*平方单位验证角度相等*)
    
40. Simplify[{4, d - e, a - e, , f - b, , (d - e)/(a - e), (c - b)/(
    
41.   f - b), , ((d - e)/(a - e))/((c - b)/(f - b)), , ((
    
42. \!\(\*OverscriptBox["d", "_"]\) -
    
43. \!\(\*OverscriptBox["e", "_"]\))/(
    
44. \!\(\*OverscriptBox["a", "_"]\) -
    
45. \!\(\*OverscriptBox["e", "_"]\)))/((
    
46. \!\(\*OverscriptBox["c", "_"]\) -
    
47. \!\(\*OverscriptBox["b", "_"]\))/(
    
48. \!\(\*OverscriptBox["f", "_"]\) -
    
49. \!\(\*OverscriptBox["b", "_"]\)))}](*验证虚部为0*)
    
50. Simplify[{5, (d - e)/(a - e)/(c - b)/(f - b), , (
    
51. \!\(\*OverscriptBox["d", "_"]\) -
    
52. \!\(\*OverscriptBox["e", "_"]\))/(
    
53. \!\(\*OverscriptBox["a", "_"]\) -
    
54. \!\(\*OverscriptBox["e", "_"]\))/(
    
55. \!\(\*OverscriptBox["c", "_"]\) -
    
56. \!\(\*OverscriptBox["b", "_"]\))/(
    
57. \!\(\*OverscriptBox["f", "_"]\) -
    
58. \!\(\*OverscriptBox["b", "_"]\))}](*验证斜线当除法是否需要加括号*)
    
59. Simplify[((d - e)/(a - e))/((c - b)/(f - b)) == ((
    
60. \!\(\*OverscriptBox["d", "_"]\) -
    
61. \!\(\*OverscriptBox["e", "_"]\))/(
    
62. \!\(\*OverscriptBox["a", "_"]\) -
    
63. \!\(\*OverscriptBox["e", "_"]\)))/((
    
64. \!\(\*OverscriptBox["c", "_"]\) -
    
65. \!\(\*OverscriptBox["b", "_"]\))/(
    
66. \!\(\*OverscriptBox["f", "_"]\) -
    
67. \!\(\*OverscriptBox["b", "_"]\)))]
    

复制代码

ccmmjj

等无人，自己做。

王守恩

\(AC=4\cos(A),CB=4\sin(A),BE=2\cos(A),EA=4-2\cos(A)\)

\(\frac{2\cos(A)}{\sin(E)}=\frac{2FE}{\sin(A)}=\frac{4-2\cos(A)}{\sin(A+E)},\frac{2\cos(A)}{\sin(A+B+E-90)}=\frac{FE}{\cos(A+B)}\)

\(已知A,\ \ 3个未知数(FE,E,B),3个方程,\ \ k=\frac{E}{B}=1\)

denglongshan

参考https://tieba.baidu.com/p/8457185129

ccmmjj126

denglongshan 发表于 2023-6-16 22:44
参考https://tieba.baidu.com/p/8457185129

刚刚去看了一下贴吧的构图，无不高深莫测。数学不能搞成机械化，也不能弄成玄学。

luyuanhong

楼上 ccmmjj 的解答已收藏。