a/ Ta có: \(\widehat{MDA}=\widehat{BDC}+\widehat{BCD}=20^o+100^o=120^o\)
\(\Rightarrow\widehat{BMN}=\widehat{DMA}=180^o-\widehat{DAM}-\widehat{ADM}=180^o-10^o-120^o=50^o\left(1\right)\)
Ta có: \(\Delta ANB\) có \(\widehat{NAB}=\widehat{NBA}=30^o\)
\(\Rightarrow\Delta ANB\) cân tại N
\(\Rightarrow NA=NB\)
Mà \(\left\{{}\begin{matrix}\widehat{CAN}=\widehat{CBN}=10^o\\CA=CB\end{matrix}\right.\)
\(\Rightarrow\Delta ACN=\Delta BCN\)
\(\Rightarrow\widehat{ACN}=\widehat{BCN}=\dfrac{100^o}{2}=50^o\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\widehat{BMN}=\widehat{BCN}\)
Mà \(\left\{{}\begin{matrix}\widehat{MBN}=\widehat{CBN}=10^o\\BN\left(chung\right)\end{matrix}\right.\)
\(\Rightarrow\Delta BMN=\Delta BCN\)
\(\Rightarrow MB=CB\)
\(\Rightarrow\Delta CBM\) cân tại B
\(\Rightarrow\widehat{BCM}=\widehat{BMC}=\dfrac{180^o-20^o}{2}=80^o\)
\(\Rightarrow\widehat{ACM}=100^o-80^o=20^o\)
b/ Theo câu a thì ta có \(\Rightarrow\Delta BMN=\Delta BCN\)
\(\Rightarrow MN=CN\left(3\right)\)
Xét \(\Delta CNE\) có
\(\left\{{}\begin{matrix}\widehat{NCE}=50^o\\\widehat{CEN}=180^o-100^o-10^o=70^o\end{matrix}\right.\)
\(\Rightarrow\widehat{CNE}=180^o-50^o-70^o=60^o\)
\(\Rightarrow\widehat{CNE}< \widehat{CEN}\)
\(\Rightarrow CE< CN\left(4\right)\)
Từ (3) và (4) \(\Rightarrow MN>CE\)
Nguyễn Huy Tú you vẽ hình đi t giải cho