O,H是△ABC的外心和垂心,AO,AH交外接圆于E,F。M,N是AC,BH中点,D=EC∩BF,证明DN⊥BM

天山草

O、H 是 △ABC 的外心和垂心，AO、AH 交 △ABC的外接圆O 于 E、F。
M、N 分别是 AC、BH 的中点。D = EC∩BF，证明 DN⊥BM。

天山草

程序代码：

1. Clear["Global`*"]; (*令△ABC的外接圆为单位圆O，BC边平行于实轴，AB、AC的复斜率分别为 u^2、v^2 *)
   
2. \!\(\*OverscriptBox[\(o\), \(_\)]\) = o = 0; a = I u v;   \!\(\*OverscriptBox[\(a\), \(_\)]\) = 1/(I u v); b = (I u)/v;  
   
3. \!\(\*OverscriptBox[\(b\), \(_\)]\) = v/(I u);  c = (I v)/u;  \!\(\*OverscriptBox[\(c\), \(_\)]\) = u/(I v);
   
4. e = -a; \!\(\*OverscriptBox[\(e\), \(_\)]\) = 1/e;
   
5. h = (I ((v^2 + 1) u^2 + v^2))/(u v); \!\(\*OverscriptBox[\(h\), \(_\)]\) = -((I (u^2 + v^2 + 1))/( u v)); (*垂心坐标*)
   
6. m = (a + c)/2; \!\(\*OverscriptBox[\(m\), \(_\)]\) = (\!\(\*OverscriptBox[\(a\), \(_\)]\) + \!\(\*OverscriptBox[\(c\), \(_\)]\))/2; n = (b + h)/2; \!\(\*OverscriptBox[\(n\), \(_\)]\) = (\!\(\*OverscriptBox[\(b\), \(_\)]\) + \!\(\*OverscriptBox[\(h\), \(_\)]\))/2;
   
7. k[a_, b_] := (a - b)/(\!\(\*OverscriptBox[\(a\), \(_\)]\) - \!\(\*OverscriptBox[\(b\), \(_\)]\)); (*复斜率定义*)
   
8. W1 = {f, \!\(\*OverscriptBox[\(f\), \(_\)]\)} /. Simplify@Solve[{(o - f) (\!\(\*OverscriptBox[\(o\), \(_\)]\) - \!\(\*OverscriptBox[\(f\), \(_\)]\)) == 1, k[a, f] == -1}, {f, \!\(\*OverscriptBox[\(f\), \(_\)]\)}] // Flatten;
   
9. f = Part[W1, 1]; \!\(\*OverscriptBox[\(f\), \(_\)]\) = Part[W1, 2];
   
10. (*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：*)
    
11. 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));
    
12. d = Simplify@Jd[k[e, c], c, k[b, f], b]; \!\(\*OverscriptBox[\(d\), \(_\)]\) = Simplify@\!\(\*OverscriptBox[\(Jd\), \(_\)]\)[k[e, c], c, k[b, f], b];
    
13. Print["M = ", Simplify[m], "， N = ", Simplify[n], "， F = ", f,  "， D = ", d];
    
14. Print["DN的复斜率kDN = ", Simplify[k[d, n]]];
    
15. Print["BM的复斜率kBM = ", Simplify[k[b, m]]];
    
16. Print["由于kDN = -kBM，所以 DN\[UpTee]BM "];

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程序运行结果：