a) Ta có:
$+ \widehat{xAy}+\widehat{x’Ay}=180^{0}$ (kề bù)
$\Rightarrow \widehat{x’Ay}=180^{0}-\widehat{xAy}=180^{0}-65^{0}=115^{0}$
$+ \widehat{xAy}+\widehat{y’Ax}=180^{0}$
$\Rightarrow \widehat{y’Ax}=180^{0}-\widehat{xAy}=180^{0}-65^{0}=115^{0}$
$+ \widehat{xAy’}+\widehat{x’Ay’}=180^{0}$
$\Rightarrow \widehat{x’Ay’}=180^{0}-\widehat{xAy’}=180^{0}-115^{0}=65^{0}$
b) Vì Az là phân giác của $\widehat{xAy}$
$\Rightarrow \widehat{xAz}=\widehat{yAz}=\frac{\widehat{xAy}}{2}=\frac{65^{0}}{2}=32,5^{0}$
Ta có: 
$+ \widehat{zAx}+\widehat{xAy’}+\widehat{z’Ay’}=180^{0}$
$\Rightarrow \widehat{z’Ay’}=180^{0}-\widehat{zAx}-\widehat{xAy’}=180^{0}-32,5^{0}-115^{0}=32,5^{0}$ (1)
$+ \widehat{zAy}+\widehat{yAx’}+\widehat{z’Ax’}=180^{0}$
$\Rightarrow \widehat{z’Ax’}=180^{0}-32,5^{0}-115^{0}=32,5^{0}$ (2)
(1)(2)$\Rightarrow Az’$ là phân giác của $\widehat{x’Ay’}$