欧拉线有关的共线
三角形ABC，BC处关于(O)的切线交于D，AE垂直于Euler line且交(O)于另一点E,DE交(O)于另一点F.设D关于BC的对称点位D'.证明：AD'F共线.
转自纯几何吧 非常适合用复平面解析几何方法做此题。
本帖最后由 dlsh 于 2023-5-10 21:03 编辑
\!\(\*OverscriptBox["o", "_"]\) = o = 0;
\!\(\*OverscriptBox["a", "_"]\) = 1/a;
\!\(\*OverscriptBox["b", "_"]\) = 1/b;
\!\(\*OverscriptBox["c", "_"]\) = 1/c; h = a + b + c;
\!\(\*OverscriptBox["h", "_"]\) =
\!\(\*OverscriptBox["a", "_"]\) +
\!\(\*OverscriptBox["b", "_"]\) +
\!\(\*OverscriptBox["c", "_"]\);(*外心在原点，并且是单位圆,H是垂心*)
d = (2 b c)/(b + c);
\!\(\*OverscriptBox["d", "_"]\) = 2/(b + c);(*D是过B和C的交点*)
KAB := (a - b)/(
\!\(\*OverscriptBox["a", "_"]\) -
\!\(\*OverscriptBox["b", "_"]\));
\!\(\*OverscriptBox["KAB", "_"]\) := 1/KAB;(*复斜率定义*)
k0 = KAB;(*欧拉直线复斜率*)
e = k0/a;
\!\(\*OverscriptBox["e", "_"]\) = 1/e;
f = -KAB/e;
\!\(\*OverscriptBox["f", "_"]\) = 1/f;
Duichengdian := (
\!\(\*OverscriptBox["a", "_"]\) b - a
\!\(\*OverscriptBox["b", "_"]\) +
\!\(\*OverscriptBox["p", "_"]\) (a - b))/(
\!\(\*OverscriptBox["a", "_"]\) -
\!\(\*OverscriptBox["b", "_"]\));
\!\(\*OverscriptBox["Duichengdian", "_"]\) := (a
\!\(\*OverscriptBox["b", "_"]\) -
\!\(\*OverscriptBox["a", "_"]\) b + p (
\!\(\*OverscriptBox["a", "_"]\) -
\!\(\*OverscriptBox["b", "_"]\)))/(a - b);(*P关于AB的对称点*)
d' = Duichengdian;
\!\(\*OverscriptBox[
RowBox[{"d", "'"}], "_"]\) =
\!\(\*OverscriptBox["Duichengdian", "_"]\);
Print["欧拉直线的复斜率=", Simplify];
Print["d=", Simplify, " \!\(\*OverscriptBox[\"d\", \"_\"]\)=",
Simplify[
\!\(\*OverscriptBox["d", "_"]\)]];
Print["e=", Simplify, " \!\(\*OverscriptBox[\"e\", \"_\"]\)=",
Simplify[
\!\(\*OverscriptBox["e", "_"]\)]];
Print["f=", Simplify, " \!\(\*OverscriptBox[\"f\", \"_\"]\)=",
Simplify[
\!\(\*OverscriptBox["f", "_"]\)]];
Print["d'=", Simplify, " \!\(\*OverscriptBox[
RowBox[{\"d\", \"'\"}], \"_\"]\)=", Simplify[
\!\(\*OverscriptBox[
RowBox[{"d", "'"}], "_"]\)]];
Print["AF的复斜率=", Simplify], " AD'的复斜率=",
Simplify]]
Print["AF和AD'的复斜率相等，所以这三点共线"]
Print["\!\(\*FractionBox[OverscriptBox[
RowBox[{\"AD\", \"'\"}], \"\\"], OverscriptBox[\"AF\", \
\"\\"]]\)=", Simplify[(d' - a)/(f - a)],
" \!\(\*OverscriptBox[
RowBox[{\" \",
RowBox[{\"(\", FractionBox[OverscriptBox[
RowBox[{\"AD\", \"'\"}], \"\\"], OverscriptBox[\"AF\", \
\"\\"]], \")\"}]}], \"_\"]\)=", Simplify[(
\!\(\*OverscriptBox[
RowBox[{"d", "'"}], "_"]\) -
\!\(\*OverscriptBox["a", "_"]\))/(
\!\(\*OverscriptBox["f", "_"]\) -
\!\(\*OverscriptBox["a", "_"]\))]]
Print["\!\(\*FractionBox[OverscriptBox[
RowBox[{\"AD\", \"'\"}], \"\\"], OverscriptBox[\"AF\", \
\"\\"]]\)=\!\(\*OverscriptBox[
RowBox[{\"(\", FractionBox[OverscriptBox[
RowBox[{\"AD\", \"'\"}], \"\\"], OverscriptBox[\"AF\", \
\"\\"]], \")\"}], \
\"_\"]\)相等，说明\!\(\*FractionBox[OverscriptBox[
RowBox[{\"AD\", \"'\"}], \"\\"], OverscriptBox[\"AF\", \
\"\\"]]\)虚部等于0，所以这三点共线"](*验证虚部等于0*)
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