a/ Ta có: \(\widehat{B_1}=\widehat{B_2}=\dfrac{\widehat{B}}{2}\)(BM phân giác \(\widehat{B}\) )
\(\widehat{C_1}=\widehat{C_2}=\widehat{\dfrac{C}{2}}\) (CN phân giác \(\widehat{C}\) )
mà
\(\widehat{B}=\widehat{C}\left(gt\right)\\ \Rightarrow\widehat{B_1}=\widehat{B_2}=\widehat{C_1}=\widehat{C_2}\)
Xét \(\Delta BNC\) và \(\Delta CMB\) có:
\(\widehat{B}=\widehat{C}\left(gt\right)\\ \widehat{C_1}=\widehat{B_1}\left(cmt\right)\)
BC cạnh chung
Vậy \(\Delta BNC=\Delta CMB\left(gcg\right)\)
\(\Rightarrow BN=CM\) (cạnh tương ứng )
b/ Ta có: \(MH\perp BC\left(gt\right)\)
mà \(NK\perp BC\left(gt\right)\)
\(\Rightarrow MH//NK\)
c/ Ta có: \(\widehat{B}+\widehat{C_1}+\widehat{BNC}=180^o\)(tổng 3 góc trong \(\Delta BNC\) )
\(\widehat{B_1}+\widehat{C}+\widehat{BMC}=180^o\)(tổng 3 góc trong \(\Delta BMC\) )
mà\(\left\{{}\begin{matrix}\widehat{B}=\widehat{C}\left(gt\right)\\\widehat{C_1}=\widehat{B}_1\left(cmt\right)\end{matrix}\right.\)
\(\Rightarrow\widehat{BNC}=\widehat{BMC}\)
Xét \(\Delta IBN\) và \(\Delta ICM\) có:
\(\widehat{B_2}=\widehat{C_2}\left(cmt\right)\\ \widehat{BNC}=\widehat{CMB}\left(cmt\right)\\ BN=MC\left(cmt\right)\)
Vậy \(\Delta IBN=\Delta ICM\left(gcg\right)\)
Chúc bạn học tốt 
