连接AD、BE, ∵BD=CE ∴弧BD=弧CE,∴∠BAD=∠EBC, ∵∠BAD=∠CAD+∠CAB,∠EBC=∠ABE+∠ABD+∠CBD, ∴∠CAD+∠CAB=∠ABE+∠ABD+∠CBD, ∵∠CAD=∠CBD(同圆中,同弧所对的圆周角相等), ∴∠CAB=∠ABD+∠ABE, ∵∠ABE=∠ACE(同圆中,同弧所对的圆周角相等), ∴∠CAB=∠ABD+∠ACE(等量代换) ∵BD、CE分别平分∠ABC、∠ACB, ∴∠ABD=∠ABC,∠ACE=∠ACB ∴∠CAB=(∠ABC+∠ACB) ∴∠ABC+∠ACB=2∠CAB ∵∠CAB+∠ABC+∠ACB=180°, ∴∠CAB+2∠CAB=180°, 3∠CAB=180° ∴∠CAB=60°. 故选C. |