ABM và
Ta có hình vẽ:
a/ Có \(\widehat{A_2}=\widehat{A_3}=90^o\left(gt\right)\Rightarrow\widehat{A_1}+\widehat{A_2}=\widehat{A_1}+\widehat{A_3}\)
hay \(\widehat{MAC}=\widehat{NAB}\)
Xét 2 t/g vuông: \(\Delta AMCvà\Delta ABN\) có:
AM = AB (gt)
\(\widehat{MAC}=\widehat{NAB}\left(cmt\right)\)
AC = AN (gt)
=> \(\Delta AMC=\Delta ABN\left(cgc\right)\left(đpcm\right)\)
b/ Vì \(\Delta AMC=\Delta ABN\left(\: ýa\right)\)
\(\Rightarrow\widehat{ACM}=\widehat{ANB}\)
Có: \(\widehat{AIN}=\widehat{CIK}\) (đối đỉnh)
\(\Delta ANI\) vuông tại A (gt)
=> \(\widehat{ANI}+\widehat{AIN}=90^o\)
mà \(\left\{{}\begin{matrix}\widehat{ACM}=\widehat{ANI}\left(cmt\right)\\\widehat{AIN}=\widehat{CIK}\left(cmt\right)\end{matrix}\right.\)
\(\Rightarrow\widehat{ANI}+\widehat{AIN}=\widehat{ACM}+\widehat{CIK}=90^o\)
Troq \(\Delta KICcó:\)
\(\widehat{IKC}+\widehat{ACM}+\widehat{CIK}=180^o\)
\(\Rightarrow\widehat{IKC}=180^o-\left(\widehat{ACM}+\widehat{CIK}\right)\)
\(=180^o-90^o=90^o\)
\(\Rightarrow MC\perp BN\left(đpcm\right)\)
c/ có r` nhé, mk k lm nx!
cảm ơn bạn nhưng mk làm x câu c rồi còn câu b thôi!
