一道难勾股定理题以△ABC的三边AB,BC,AC为边长向形外作正△ACF,正△ABE,正△BCD,若S△ACF+S△ABE=S△BCD,证明:△ABC为直角△.

Question

一道难勾股定理题
以△ABC的三边AB,BC,AC为边长向形外作正△ACF,正△ABE,正△BCD,若S△ACF+S△ABE=S△BCD,证明:△ABC为直角△.

Answer 1

设△ABC边长分别为AB=a,BC=b,CA=c.
S△ACF=c/2*c*sin60
S△ABE=a/2*a*sin60
S△BCD=b/2*b*sin60
S△ACF+S△ABE=（c/2*c+a/2*a）*sin60
有,b^2=a^2+c^2
得：△ABC为直角△.