+) Ta có :
\(AMC+CMB=180^0\) (kề bù)
Mà \(BMC=3.CMA\)
\(\Leftrightarrow CMA+3CMA=180^0\)
\(\Leftrightarrow CMA.\left(1+3\right)=180^0\)
\(\Leftrightarrow CMA.4=180^0\)
\(\Leftrightarrow CMA=45^0\)
\(\Leftrightarrow BMC=135^0\)
+) Ta có :
\(AMC=BMD\) (đối đỉnh)
Mà \(AMC=45^0\)
\(\Leftrightarrow BMD=45^0\)
+) Ta có :
\(BMC=AMD\) (đối đỉnh)
Mà \(BMC=135^0\)
\(\Leftrightarrow AMD=135^0\)
Ta có: \(\widehat{BMC}+\widehat{CMA}=180^o\)(kề bù)
mà \(\widehat{BMC}=3\widehat{CMA}\) (gt)
\(\Rightarrow3\widehat{CMA}+\widehat{CMA}=180^o\)
\(\Rightarrow4\widehat{CMA}=180\Rightarrow\widehat{CMA}=45^o\)
\(\widehat{BMC}=3\widehat{CMA}=3.45^o=135^o\)
\(\widehat{DMB}=\widehat{CMA}=45^o\)(đối đỉnh)
\(\widehat{AMD}=\widehat{BMC}=135^o\)(đối đỉnh)
Vậy ...
Ta có: \(\widehat{BMC}+\widehat{CMA}=180^0\)
\(\Leftrightarrow3\widehat{CMA}+\widehat{CMA}=180^0\)
\(\Leftrightarrow4\widehat{CMA}=180^0\)
\(\Leftrightarrow\widehat{CMA}=180^0:4=45^0\)
\(\Rightarrow\widehat{BMC=}180^0-45^0=135^0\)
Ta có: \(\widehat{CMA}=\widehat{BMD}\)(đối đỉnh)
\(\Rightarrow\widehat{BMD}=45^0\)
\(\widehat{BMC}=\widehat{AMD}\)( đối đỉnh)
\(\Rightarrow\widehat{AMD}=135^0\)
