凸四边形ADBC内接于圆M,AB是圆的直径.E,F分别是AC,BD的中点.....,证明NTPQ四点共圆。

天山草

凸四边形 ADBC 内接于圆M，AB 是圆的直径。E、F 分别是 AC、BD 的中点。
H = BC∩FE，G = DA∩FE，以 EH 为直径的圆O1 和以 FG 为直径的圆O2 交于 P、Q 两点。
T = CD∩AB，N = QM∩圆MDC。证明 NTPQ 四点共圆。

天山草

程序代码：

1. Clear["Global`*"];
   
2. \!\(\*OverscriptBox[\(m\), \(_\)]\) = m = 0; \!\(\*OverscriptBox[\(a\), \(_\)]\) = a = -1;
   
3. \!\(\*OverscriptBox[\(b\), \(_\)]\) = b = 1; kAC = u^2; kBC = -u^2; \!\(\*OverscriptBox[\(t\), \(_\)]\) = t;
   
4. k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
   
5. (*过A1点、复斜率等于k1的直线，与过A2点、复斜率等于k2的直线的交点：*)
   
6. Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\)) - k1 (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k1_, a1_, k2_, a2_] := -((a1 - k1 \!\(\*OverscriptBox[\(a1\), \(_\)]\) - (a2 - k2 \!\(\*OverscriptBox[\(a2\), \(_\)]\)))/(k1 - k2));
   
7. c = Simplify@Jd[kAC, a, kBC, b]; \!\(\*OverscriptBox[\(c\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[kAC, a, kBC, b];
   
8. W1 = {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)} /. Simplify@Solve[{(m - d) (\!\(\*OverscriptBox[\(m\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) == 1, k[c, d] == k[c, t]}, {d, \!\(\*OverscriptBox[\(d\), \(_\)]\)}] // Flatten;
   
9. d = Part[W1, 1]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = Part[W1, 2];
   
10. e = (a + c)/2; \!\(\*OverscriptBox[\(e\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(c\), \(_\)]\))/2; f = (b + d)/2; \!\(\*OverscriptBox[\(f\), \(_\)]\) = (\!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(d\), \(_\)]\))/2;
    
11. h = Simplify@Jd[k[b, c], c, k[f, e], e]; \!\(\*OverscriptBox[\(h\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k[b, c], c, k[f, e], e];
    
12. g = Simplify@Jd[k[d, a], a, k[f, e], e]; \!\(\*OverscriptBox[\(g\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k[d, a], a, k[f, e], e];
    
13. o1 = (e + h)/2; \!\(\*OverscriptBox[\(o1\), \(_\)]\) = (\!\(\*OverscriptBox[\(e\), \(_\)]\) + \!\(\*OverscriptBox[\(h\), \(_\)]\))/2; o2 = (g + f)/2; \!\(\*OverscriptBox[\(o2\), \(_\)]\) = (\!\(\*OverscriptBox[\(g\), \(_\)]\) + \!\(\*OverscriptBox[\(f\), \(_\)]\))/2;
    
14. W2 = {z, \!\(\*OverscriptBox[\(z\), \(_\)]\)} /. Simplify@Solve[{(o1 - c) (\!\(\*OverscriptBox[\(o1\), \(_\)]\) - \!\(\*OverscriptBox[\(c\), \(_\)]\)) == (o1 - z) (\!\(\*OverscriptBox[\(o1\), \(_\)]\) - \!\(\*OverscriptBox[\(z\), \(_\)]\)), (o2 - d) (\!\(\*OverscriptBox[\(o2\), \(_\)]\) - \!\(\*OverscriptBox[\(d\), \(_\)]\)) == (o2 - z) (\!\(\*OverscriptBox[\(o2\), \(_\)]\) - \!\(\*OverscriptBox[\(z\), \(_\)]\))}, {z, \!\(\*OverscriptBox[\(z\), \(_\)]\)}] // Flatten;
    
15. p = Part[W2, 1]; \!\(\*OverscriptBox[\(p\), \(_\)]\) = Part[W2, 2]; q = Part[W2, 3]; \!\(\*OverscriptBox[\(q\), \(_\)]\) = Part[W2, 4];
    
16. WX[a_, b_, c_] := (a \!\(\*OverscriptBox[\(a\), \(_\)]\) (c - b) + b \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - c) + c \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a) )/(\!\(\*OverscriptBox[\(a\), \(_\)]\) (c - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a));(*三角形 ABC 的外心坐标：*)   
    
17. \!\(\*OverscriptBox[\(WX\), \(_\)]\)[a_, b_, c_] := (\!\(\*OverscriptBox[\(a\), \(_\)]\) \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\) \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) \!\(\*OverscriptBox[\(a\), \(_\)]\) (c - a))/(\!\(\*OverscriptBox[\(a\), \(_\)]\) (c - b) + \!\(\*OverscriptBox[\(b\), \(_\)]\) (a - c) + \!\(\*OverscriptBox[\(c\), \(_\)]\) (b - a));
    
18. o3 = Simplify@WX[d, c, m]; \!\(\*OverscriptBox[\(o3\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(WX\), \(_\)]\)[d, c, m];
    
19. Print["c = ", c, "，  d = ", d, "，  h = ", h, "，  g = ", g];
    
20. W3 = {n, \!\(\*OverscriptBox[\(n\), \(_\)]\)} /. Simplify@Solve[{(o3 - m) (\!\(\*OverscriptBox[\(o3\), \(_\)]\) - \!\(\*OverscriptBox[\(m\), \(_\)]\)) == (o3 - n) (\!\(\*OverscriptBox[\(o3\), \(_\)]\) - \!\(\*OverscriptBox[\(n\), \(_\)]\)), k[q, m] == k[q, n], n != m}, {n,
    
21. \!\(\*OverscriptBox[\(n\), \(_\)]\)}] // Flatten;
    
22. n = Part[W3, 1]; \!\(\*OverscriptBox[\(n\), \(_\)]\) = Part[W3, 2];
    
23. Print["o3 = ", o3, "，  p = ", p, "，  q = ", q, "，  n = ", n];
    
24. Print["复斜率 kQT/kQP = ", Simplify[k[q, t]/k[q, p]]];
    
25. Print["复斜率 kNT/kNP = ", Simplify[k[n, t]/k[n, p]]];
    
26. Print["由于 kQT/kQP = kNT/kNP，所以 \[Angle]TQP = \[Angle]TNP，故 NTPQ 四点共圆 "];

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程序运行结果：