Giải:
a) Ta có: \(\widehat{ABC}=\widehat{ACB}\) ( t/g ABC cân tại A )
\(\Rightarrow\widehat{ABC}-60^o=\widehat{ACB}-60^o\)
\(\Rightarrow\widehat{ABC}-\widehat{DBC}=\widehat{ACB}-\widehat{DCB}\)
\(\Rightarrow\widehat{ABD}=\widehat{ACD}\)
Xét \(\Delta ABD,\Delta ACD\) có:
\(AB=AC\) ( t/g ABC cân tại A )
\(\widehat{ABD}=\widehat{ACD}\) ( cmt )
BD = DC ( t/g DBC đều )
\(\Rightarrow\Delta ABD=\Delta ACD\left(c-g-c\right)\)
\(\Rightarrow\widehat{BAD}=\widehat{CAD}\) ( góc t/ứng )
\(\Rightarrow\)AD là phân giác của \(\widehat{BAC}\) ( đpcm )
b) Ta có: \(\widehat{ABC}=\widehat{ACB}=80^o\)
\(\Rightarrow\widehat{ABD}=\widehat{ABC}-\widehat{DBC}=80^o-60^o=20^o\)
\(\Rightarrow\widehat{ABM}=\dfrac{1}{2}\widehat{ABD}1=10^o\)
Xét \(\Delta ABM,\Delta BAD\) có:
\(\widehat{BAM}=\widehat{ABD}=20^o\)
AB: cạnh chung
\(\widehat{ABM}=\widehat{BAD}=10^o\)
\(\Rightarrow\Delta ABM=\Delta BAD\left(g-c-g\right)\)
\(\Rightarrow AM=BD\) ( cạnh t/ứng )
Mà BD = BC ( t/g DBC đều )
\(\Rightarrow AM=BC\left(đpcm\right)\)
Vậy...
