$\widehat{A_1}=50^o$
$⇒\widehat{A_1}=\widehat{A_2}=50^o$ (đối đỉnh)
$xx'//yy'$
$⇒\widehat{A_2}+\widehat{ABy'}=180^o$ (trong cùng phía bù nhau)
mà $\widehat{A_2}=50^o$
$⇒\widehat{ABy'}=180^o-50^o=130^o$
mà $Bm$ là phân giác $\widehat{ABy'}$
$⇒\widehat{B_4}=\dfrac{\widehat{ABy'}}{2}=\dfrac{130^o}{2}=65^o$
$M∈xx'$
$⇒Mx'//By'$
$⇒\widehat{M_6}+\widehat{B_4}=180^o$
mà $\widehat{B_4}=65^o$
$⇒\widehat{M_6}=180^o-65^o=115^o$
Vậy $\widehat{M_6}=115^o$