$\text{Kẻ Mx//PN ta có :}$
$\text{$\widehat{M_2}$=$180^o$-$\widehat{N}$(Trong cùng phía bù nhau)}$
$\text{⇒$\widehat{M_2}$=$180^o$-$90^o$=$90^o$}$
$\text{⇒$\widehat{M_1}$=$135^o$-$90^o$=$45^o$}$
$\text{Mà $\widehat{ABM}$+$\widehat{M_1}$=$45^o$ + $135^o$=$180^o$(Trong cùng phía)}$
$\text{⇒AB//Mx}$
$\text{⇒AB//Mx//PN (Tính chất bắc cầu)}$