在三角形abc中,(sina+sinb+sinc)^2=3(sin^2a+sin^2b+sin^2c) 求三角形形状

(sinA)^2＋(sinB)^2＋(sinC)^2＋2sinAsinB＋2sinBsinA＋2sinCsinA
＝3(sinA)^2＋3(sinB)^2＋3(sinC)^2
2(sinA)^2＋2(sinB)^2＋2(sinC)^2－2sinAsinB－2sinBsinA－2sinCsinA＝0
[(sinA)^2＋(sinB)^2－2sinAsinB]＋[(sinB)^2＋(sinC)^2－2sinBsinC]＋[(sinC)^2＋(sinA)^2－2sinCsinA]＝0
(sinA－sinB)^2＋(sinB－sinC)^2＋(sinC－sinA)^2＝0
∴sinA＝sinB＝sinC
∴∠A＝∠B＝∠C
等边三角形