Ta có:
\(\left\{{}\begin{matrix}BM\perp AC\left(gt\right)\\CN\perp AB\left(gt\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\widehat{AMB}=90^0\\\widehat{ANC}=90^0\end{matrix}\right.\)
+ Xét \(\Delta ABM\) có:
\(\widehat{A}+\widehat{ABM}+\widehat{AMB}=180^0\) (định lí tổng 3 góc trong một tam giác).
=> \(\widehat{A}+\widehat{ABM}+90^0=180^0\)
=> \(\widehat{A}+\widehat{ABM}=180^0-90^0\)
=> \(\widehat{A}+\widehat{ABM}=90^0\) (1).
+ Xét \(\Delta ACN\) có:
\(\widehat{A}+\widehat{ACN}+\widehat{ANC}=180^0\) (như ở trên).
=> \(\widehat{A}+\widehat{ACN}+90^0=180^0\)
=> \(\widehat{A}+\widehat{ACN}=180^0-90^0\)
=> \(\widehat{A}+\widehat{ACN}=90^0\) (2).
Từ (1) và (2) \(\Rightarrow\widehat{A}+\widehat{ABM}=\widehat{A}+\widehat{ACN}.\)
=> \(\widehat{ABM}=\widehat{ACN}\left(đpcm\right).\)
Chúc bạn học tốt!
ABC cân tại A, góc A = 500:
ABH=
ACH.
