Question

已知三角形ABC中CD,CE分别是高和中线,且角ACE=角ECD=角DCB,求证角ACD=90度

Answer 1

CD⊥AB,且∠ACE=∠ECD=∠DCB则：ED=DB=EB/2=AE/2由角平分线定理：AC/CD=AE/ED=2/1即：CD=AC/2,而CD⊥AB可知：∠A=30°也可知：∠ACE=∠ECD=∠DCB=(90°-∠A)/2=(90°-30°)/2=30°所以：∠ACB=∠ACE+∠ECD+∠DCB=90...