a) $xy$ // x' y'x′y′ nên \widehat{xAB}=\widehat{ABy'}xAB=ABy′​ (hai góc so le trong). (1)

{AA}'AA′ là tia phân giác của \widehat{xAB}xAB nên: \widehat{{A}_{1}}=\widehat{{A}_{2}}=\dfrac{1}{2} \widehat{{xAB}}A1​​=A2​​=21​xAB (2)

{BB}'BB′ là tia phân giác của \widehat{{ABy}'}ABy′​ nên: \widehat{B_{1}}=\widehat{B_{2}}=\dfrac{1}{2} \widehat{A B y'}B1​

a) $xy$ // x' y'x′y′ nên \widehat{xAB}=\widehat{ABy'}xAB=ABy′​ (hai góc so le trong). (1)

{AA}'AA′ là tia phân giác của \widehat{xAB}xAB nên: \widehat{{A}_{1}}=\widehat{{A}_{2}}=\dfrac{1}{2} \widehat{{xAB}}A1​​=A2​​=21​xAB (2)

{BB}'BB′ là tia phân giác của \widehat{{ABy}'}ABy′​ nên: \widehat{B_{1}}=\widehat{B_{2}}=\dfrac{1}{2} \widehat{A B y'}B1​

a) $xy$ // x' y'x′y′ nên \widehat{xAB}=\widehat{ABy'}xAB=ABy′​ (hai góc so le trong). (1)

{AA}'AA′ là tia phân giác của \widehat{xAB}xAB nên: \widehat{{A}_{1}}=\widehat{{A}_{2}}=\dfrac{1}{2} \widehat{{xAB}}A1​​=A2​​=21​xAB (2)

{BB}'BB′ là tia phân giác của \widehat{{ABy}'}ABy′​ nên: \widehat{B_{1}}=\widehat{B_{2}}=\dfrac{1}{2} \widehat{A B y'}B1​

a) \(x y\) // \(x^{'} y^{'}\) nên \(\hat{x A B} = \hat{A B y^{'}}\) (hai góc so le trong). (1)

\(\left(A A\right)^{'}\) là tia phân giác của \(\hat{x A B}\) nên: \(\hat{A_{1}} = \hat{A_{2}} = \frac{1}{2} \hat{x A B}\) (2)

\(\left(B B\right)^{'}\) là tia phân giác của \(\hat{\left(A B y\right)^{'}}\) nên: \(\hat{B_{1}} = \hat{B_{2}} = \frac{1}{2} \hat{A B y^{'}}\) (3)

Từ (1), (2), (3) ta có: \(\hat{A_{2}} = \hat{B_{1}}\).

Mà hai góc ở vị trí so le trong, nên \(\left(A A\right)^{'} / / \left(B B\right)^{'}\)

b) \(x y\) // \(x^{'} y^{'}\) nên \(\hat{A_{1}} = \hat{\left(A A\right)^{'} B}\) (hai góc so le trong).

\(\left(A A\right)^{'} / / \left(B B\right)^{'}\) nên \(\hat{A_{1}} = \hat{\left(A B\right)^{'} B}\) (hai góc đồng vị).

Vậy \(\hat{\left(A A\right)^{'} B} = \hat{\left(A B\right)^{'} B}\).

AA'//BB'

50

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