Chứng minh BN vuông góc CM biết tam giác ABC có góc A nhỏ hơn 90 độ, AH vuông góc BC

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!