blue bigbang微盘:“拿破仑三角形”的非几何证明，比如三角函数方法。

嗯，关注中.....下午找时间看看，留个记号，省得翻页了，呵呵
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应用余弦定理来来证明三边相等，思路是对的，再进一步配合正弦定理不就可以了吗？哈哈哈....

关键点：证明边相等不需要求出边长，只要边长之差为 0 或者边长之比为 1 就可以了。

先画图：△ABC，按照习惯设顶点相对边长分别为a、b、c，BC边外正三角形中心X、CA边为Y、AB边为Z，同样设△XYZ顶点相对边长为x、y、z

考察 △CYA 有：AY = CY = b/√3，同样，在△AZB 中 AZ = c/√3，其他类推....应用余弦定理得到：

x^2 = [b^2 + c^2 - 2bc*cos(A + π/3)]/3 ............... (1)
y^2 = [c^2 + a^2 - 2ca*cos(B + π/3)]/3 ............... (2)
z^2 = [a^2 + b^2 - 2ab*cos(C + π/3)]/3 ............... (3)

此时应用正弦定理得到关系式：(sinA)/a = (sinB)/b = (sinC)/c = k>0 其中k是比例常数，无需知道具体数值，代入上面三式，可以得到——

3(kx)^2 = (sinB)^2 + (sinC)^2 - 2sinB*sinC*cos(A + π/3) ........ (4)
3(ky)^2 = (sinC)^2 + (sinA)^2 - 2sinC*sinA*cos(B + π/3) ........ (5)
3(kz)^2 = (sinA)^2 + (sinB)^2 - 2sinA*sinB*cos(C + π/3) ........ (6)

(4) - (5)得到

3k^2*(x - y)^2
= [(sinB)^2 - (sinA)^2] + 2sinC*[sinA*cos(B + π/3) - sinB*cos(A + π/3)]
= [cos2A - cos2B]/2 + sinC*[sin(A + B + π/3) + sin(A - B - π/3) - sin(A + B + π/3) -sin(B - A - π/3)]
= [cos2A - cos2B]/2 + sinC*[sin(A - B - π/3) + sin(A - B + π/3)]
= [cos2A - cos2B]/2 + sinC*2sin(A - B)*cos(2π/3) .............. cos(2π/3) = -1/2
= [cos2A - cos2B]/2 - sinC*sin(A - B)
= sin(A + B)*sin(A - B) - sinC*sin(A - B)

注意，C = π - (A + B)，sinC = sin(A + B)

3k^2*(x - y)^2 = 0 ..... x = y，同样可以得到 y = z

△XYZ为等边三角形，证毕。