Ta có: \(\left\{{}\begin{matrix}Bx\perp BC\left(gt\right)\Rightarrow\widehat{CBx}=90^0\\Cy\perp BC\left(gt\right)\Rightarrow\widehat{BCy}=90^0\end{matrix}\right.\)
Lại có:
\(\widehat{CBx}=\widehat{ABx}+\widehat{ABC}=90^0\) (1)
\(\widehat{BCy}=\widehat{ACB}+\widehat{ACy}=90^0\) (2)
Từ (1) và (2) => \(\widehat{ABx}+\widehat{ABC}+\widehat{ACB}+\widehat{ACy}=90^0+90^0\)
=> \(\widehat{ABx}+\widehat{ABC}+\widehat{ACB}+\widehat{ACy}=180^0\) (3)
Xét \(\Delta ABC\) có \(\widehat{A}=90^0\left(gt\right)\)
=> \(\widehat{ABC}+\widehat{ACB}=90^0\) (tính chất tam giác vuông)
Thay vào (3) ta được:
\(\widehat{ABx}+\widehat{ACy}+90^0=180^0\)
=> \(\widehat{ABx}+\widehat{ACy}=180^0-90^0\)
=> \(\widehat{ABx}+\widehat{ACy}=90^0.\)
Chúc bạn học tốt!
