Giải:
Ta có:
\(A\widehat{M}C+C\widehat{M}B=180^o\)
\(\Leftrightarrow3B\widehat{M}C+C\widehat{M}B=180^o\)
\(\Leftrightarrow4B\widehat{M}C=180^o\)
\(\Leftrightarrow B\widehat{M}C=45^o\)
\(A\widehat{M}C=3B\widehat{M}C=3.45^o=135^o\)
\(A\widehat{M}C+A\widehat{M}D=180^o\)
\(\Leftrightarrow A\widehat{M}D=180^o-A\widehat{M}C=180^o-135^o=45^o\)
\(B\widehat{M}D+B\widehat{M}C=180^o\)
\(\Leftrightarrow B\widehat{M}D=180^o-B\widehat{M}C=180^o-45^o=135^o\)
Vậy \(A\widehat{M}C=B\widehat{M}D=135^o;C\widehat{M}B=A\widehat{M}D=45^o\)
Đúng 1
Bình luận (0)
