The bisectors are  

$\begin{array}{r}\frac{2x-y+4}{\sqrt{5}}=\pm \frac{(x-2y-1)}{\sqrt{5}}\\ \Rightarrow x-y+1=0,x+y+5=0\end{array}$

We take any one given line and any one angular bisector..

Consider $2x-y+4=0\text{ and }x-y+1=0$ with slopes 

2 and 1 .

The angle between them is 

${\tan }^{-1}\left(\frac{2-1}{1+2}\right)={\tan }^{-1}\frac{1}{3}<\frac{\pi }{4}$

$\therefore$The acute angle bisector is $x-y+1=0$

The obtuse angle bisector is the other bisector

$x+y+5=0$.