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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

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发表于 2023-10-27 18:50 | 显示全部楼层 |阅读模式
本帖最后由 天山草 于 2023-10-27 19:03 编辑

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。

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 楼主| 发表于 2023-10-27 18:54 | 显示全部楼层
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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