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O A B x y x' y'
góc AOy + góc OAy' = 180 độ (xy//x'y') (1)
góc AOB = góc AOy : 2 (OB là tia phân giác của góc AOy) (2)
góc OAB = góc OAy' : 2 (AB là tia phân giác của góc OAy') (3)
Từ (1); (2); (3) => góc AOB + góc OAB = (góc AOy + góc OAy') : 2 = 180 độ : 2 = 90 độ
=> tam giác OAB vuông tại B (DHNB)
=> OB vuông góc với AB (t/c)
a b A B C 1 2 1 2 c
a // b
c x a = A
c x b = B
\(\begin{cases}\widehat{A_1}=\widehat{A_2}=\frac{1}{2}.\widehat{A}\\\widehat{B_1}=\widehat{B_2}=\frac{1}{2}.\widehat{B}\end{cases}\)
Mặt khác
\(\widehat{A}+\widehat{B}=180^0\)
=> \(\widehat{A_1}+\widehat{B_1}=\frac{\widehat{A}}{2}+\frac{\widehat{B}}{2}\)
=> \(\widehat{A_1}+\widehat{B_1}=\frac{\widehat{A}+\widehat{B}}{2}\)
=> \(\widehat{A_1}+\widehat{B_1}=\frac{180^0}{2}=90^0\)
Xét \(\Delta ABC\) có :
\(\widehat{A_1}+\widehat{B_1}+\widehat{C}=180^0\)
=> \(90^0+\widehat{C}=180^0\)
=> \(\widehat{C}=90^0\) ( đpcm )
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A B x y C 1 1
\(\widehat{xAB}+\widehat{yBA}=180^0\)(2 góc trong cùng phía của Ax // By)
mà\(\widehat{A_1}=\frac{\widehat{xAB}}{2};\widehat{B_1}=\frac{\widehat{yBA}}{2}\)(AC,BC là phân giác của\(\widehat{xAB};\widehat{yBA}\))
=>\(\Delta ABC\)có :\(\widehat{A_1}+\widehat{B_1}=\frac{\widehat{xAB}+\widehat{yBA}}{2}=\frac{180^0}{2}=90^0\)\(\Rightarrow\Delta ABC\)vuông tại C hay AC _|_ BC
a b N 1 1 2 2 U R T z m
Ta có :
Góc RUN = Góc UNT ( so le trong )
\(\Rightarrow\)\(\frac{1}{2}\)Góc RUN = \(\frac{1}{2}\)Góc UNT
\(\Rightarrow\)Góc U1 = Góc N1 ( =\(\frac{1}{2}\)Góc RUN = \(\frac{1}{2}\)Góc UNT )
Mà đây là 2 góc so le trong
\(\Rightarrow\)Uz // Nm ( theo dấu hiệu nhiên biết 2 đường thắng song song )
\(\Rightarrow DPCM\)
Vậy ...
Vì hai đường song song thì có hai góc cùng phía bù nhau
=> Tổng hai góc cùng phía = 1800
=> Tổng hai góc phân giác của hai góc cùng phía = 900
=> Hai tia phân giác của hai góc trong cùng phía là góc vuông (ĐPCM)
mù ak, ghi dấu rùi ây! ko tl dc thì ra chỗ khác đỡ tốn chỗ giải toán
A B O a b 1 2 1 2 C D
Cho hình vẽ như trên.
Ta có:
a//b => góc CAB + góc ABD = 1800 (trong cùng phía)
Mà Â1= Â2, góc B1 góc B2
Nên 2.Â2 + 2. góc B2 = 1800
=> Â2 + góc B1 = 900
Tam giác AOB có:
Â2 + góc B1 + AÔB =1800
Hay AÔb = 1800 - (Â2 + góc B1) = 1800 - 900 = 900
=>OA vuông góc với OB (ĐPCM)