再来一题，详情进贴

∵∠ADF+∠BDE=120°；∠BDE+∠BED=120°
∴∠ADF=∠BED
∴△ADF≌△BED
∴AF=BD
∴AC=AB
同理可证：AC=BC
∴△ABC是正三角形

cos(π/13)+cos(3π/13)+cos(9π/13)
令a= cos(π/13)+cos(3π/13)+cos(9π/13)
∵ cos(π/13)＞0，cos(3π/13)+cos(9π/13)=cos(3π/13)-cos(4π/13)＞0
∴a＞0
令b= cos(5π/13)+cos(7π/13)+cos(11π/13)
∵2sin(π/13)*(a+b)
=2sin(π/13)*[cos(π/13)+cos(3π/13)+cos(5π/13)+ cos(7π/13)+cos(9π/13)+cos(11π/13)]
=sin(2π/13)+sin(4π/13)-sin(2π/13)+sin(6π/13)-sin(4π/13)++sin(8π/13)-sin(6π/13)+sin(10π/13)-sin(8π/13)+sin(12π/13)-sin(10π/13)
=sin(12π/13)
=sin(π/13)
∴a+b=1/2
∵2ab=2*[cos(π/13)+cos(3π/13)+cos(9π/13)]*[cos(5π/13)+cos(7π/13)+cos(11π/13)]
=cos(4π/13)+cos(6π/13)+cos(6π/13)+cos(8π/13)+cos(10π/13)+cos(12π/13)+cos(2π/13)+cos(8π/13)+cos(4π/13)+cos(10π/13)+cos(8π/13)+cos(14π/13)+cos(4π/13)+cos(14π/13)+cos(2π/13)+cos(16π/13)+cos(2π/13)+cos(20π/13)
=-3*[cos(π/13)+cos(3π/13)+cos(5π/13)+ cos(7π/13)+cos(9π/13)+cos(11π/13)]
=-3/2
∴ab=-3/4
所以a是方程x²-x/2-3/4=0的正根，即：a=(1+√13)/4

图一，因为AD=BE=CF,又三角形ABC为正三角形，所以AF=BD=EC，角A,B,C相等，易证三个钝角三角形全等，所以中间三角形三边相等，为等边三角形。手机写数学语言麻烦，见谅。

ED=EF=FD正三角形相等
AD=BE=CF
DBE=DAF=FEC
AB=BC=CA
ABC是正三角形

第一题也太简单了吧。。。。

第一题， 因为DFE是正三角形 所以三条边相等，且每个角都是60°，所以DF,FE,ED,三条边相等 ，又因为AD,BE,CF,相等 所以三角形ADF,三角形CFE,三角形BED相等，又因为三角形ADF三角形CFE三角形BED三个三角形相等和AD=BE=CF
所以AD+DB=BE+EC=CF+EA，三条边相等，所以AB=BC=CA,已知三条边相等所以三角形ABC是正三角形，