Gọi O à 1 điểm nằm trên đường trung trực của BC (O thuộc BC)
Xét \(\Delta ABM\)và \(\Delta OBM\)có
\(\widehat{ABM}=\widehat{MBO}\)(gt)
BM chung
\(\widehat{A}=\widehat{BOM}\)(=90o)
=> \(\Delta ABM\)=\(\Delta OBM\)(ch-gn)
=> \(\widehat{AMB}=\widehat{BMO}\)(cặp góc tương ứng)
Xét\(\Delta MBO\)và\(\Delta MCO\) có
MO chung
\(\widehat{MOB}=\widehat{MOC}\)(=900)
BO=OC
=> \(\Delta MBO\)=\(\Delta MCO\)(2cgv)
=>\(\widehat{BMO}=\widehat{CMO}\)(cgtư)
.=> \(\widehat{AMB}=\widehat{BMO}\)=\(\widehat{CMO}\)
mà \(\widehat{AMB}+\widehat{BMO}+\widehat{CMO}=180^o\)
=>\(\widehat{AMB}=\widehat{BMO}=\widehat{CMO}=60^0\)
=> \(\widehat{ACB}=90^{o^{ }}-60^0=30^0\)