AC = HC BC2AB HB⇒ =2AC HC2 4HB AB⇒ =2 4HC ACLẠI CÓ
2
.
AC = HC BC
2
AB HB⇒ =AC HC2
4
HB ABHC ACLại có:
HB BE BA
2
= .
HC CF CA
2
= .
HB BE AB.HC CF AC4
AB BE ABAC CF AC3
AB BE⇔ =AC CF2
2
AB HB⇒ =AC HC2
4
HB ABHC ACLại có:
2
2
4
AB BE ABAC CF AC3
AB BE⇔ =AC CF