Lời giải:
a. Vì $xx'\parallel yy'$ nên $\widehat{x'AB}+\widehat{ABy'}=180^0$ (hai góc trong cùng phía)
Mà:
$\widehat{x'AB}=2\widehat{mAB}$
$\widehat{ABy'}=2\widehat{ABn}$
$\Rightarrow 2(\widehat{mAB}+\widehat{ABn})=180^0$
$\Rightarrow \widehat{mAB}+\widehat{ABn}=90^0$
Gọi giao điểm cả $Am, Bn$ là $C$ thì:
$\widehat{CAB}+\widehat{ABC}=90^0$
$\Rightarrow \widehat{AcB}=180^0-(\widehat{CAB}+\widehat{ABC})=180^0-90^0=90^0$
$\Rightarrow AM\perp Bn$
b.
$\widehat{xAz}=\frac{1}{2}\widehat{xAB}$
$\widehat{nBy'}=\frac{1}{2}\widehat{ABy'}$
Mà $\widehat{xAB}=\widehat{ABy'}$ (hai góc so le trong với $xx'\parallel yy'$)
$\Rightarrow \widehat{xAz}=\widehat{nBy'}$
