ABCD 是正方形，E,F 在 BC,AB 上，EF=ED，P 在 BE 上，∠PFE=∠EDC。证：∠BPF=2∠CDP

kanyikan

denglongshan

1. 
   
2. 
   
3. \!\(\*OverscriptBox["b", "_"]\) = b = 0;
   
4. \!\(\*OverscriptBox["c", "_"]\) = c = 1; d = 1 + I;
   
5. \!\(\*OverscriptBox["d", "_"]\) = 1 - I; a = I;
   
6. \!\(\*OverscriptBox["a", "_"]\) = -I;
   
7. \!\(\*OverscriptBox["f", "_"]\) = -f;
   
8. m = (d + f)/2;
   
9. \!\(\*OverscriptBox["m", "_"]\) = (
   
10. \!\(\*OverscriptBox["d", "_"]\) +
    
11. \!\(\*OverscriptBox["f", "_"]\))/2;(*M是DF中点*)
    
12. k[a_, b_] := (a - b)/(
    
13. \!\(\*OverscriptBox["a", "_"]\) -
    
14. \!\(\*OverscriptBox["b", "_"]\));
    
15. \!\(\*OverscriptBox["k", "_"]\)[a_, b_] := 1/k[a, b];(*复斜率定义*)
    
16. 
    
17. \!\(\*OverscriptBox["Jd", "_"]\)[k1_, a1_, k2_, a2_] := -((a1 - k1
    
18. \!\(\*OverscriptBox["a1", "_"]\) - (a2 - k2
    
19. \!\(\*OverscriptBox["a2", "_"]\)))/(
    
20.   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*)
    
21. Jd[k1_, a1_, k2_, a2_] := -((k2 (a1 - k1
    
22. \!\(\*OverscriptBox["a1", "_"]\)) - k1 (a2 - k2
    
23. \!\(\*OverscriptBox["a2", "_"]\)))/(k1 - k2));
    
24. e = Jd[-k[d, f], m, 1, b];
    
25. \!\(\*OverscriptBox["e", "_"]\) =
    
26. \!\(\*OverscriptBox["Jd", "_"]\)[-k[d, f], m, 1, b];
    
27. kPF = -k[f, e] k[d, e];(*角相等条件*)
    
28. p = Jd[kPF, f, 1, b];
    
29. \!\(\*OverscriptBox["p", "_"]\) =
    
30. \!\(\*OverscriptBox["Jd", "_"]\)[kPF, f, 1, b];
    
31. Simplify[{e,
    
32. \!\(\*OverscriptBox["e", "_"]\)}]
    
33. Simplify[{1/k[p, f], -1/k[p, d], (-1/k[p, d])^2}](*验证结论*)
    

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luyuanhong

王守恩

\(记AB=1,\ \ \ ∠ADF=a(已知条件),\ \ \ ∠PDE=b,\ \ \ ∠EDC=c,\ \ \ 解方程\)

\((1-\tan(a))^2+(1-\tan(c))^2=(\frac{1}{\cos(c)})^2\ \ \ \ \ \ (1-\tan(a))\tan(2a)+\tan(b+c)=1\)

\(可得\ ∠BPF=2∠CDP\ \ \ \ \ \ 兼得∠PDF=45^\circ,\ \ \ \ PD是∠FPC的角平分线。\)

kanyikan

luyuanhong

楼上 kanyikan 的解答很好！已收藏。

litaolitt

直接延长至ad,(ef,pf),做点p,e到ad的垂直线