Special angle?

Considering that angle $A$ is double the size of angle $B$ , we can bisect angle $A$ like we did for angle $C$ . Now suppose the angle bisector of angle $A$ meets $BC$ at point $E$ such that angle $ABC$ = angle $BAE$ = angle $CAE=x$ where $x$ is the unknown angle. Considering that angle $ABE$ =angle $BAE$ , this tells us that $AE=BE$ . Next join point $D$ to $E$ :

Considering that $AC=BD$ as given in the question, $AE=BE$ as determined and angle $CAE$ = angle $DBE$ , we see that triangles $ACE$ and $BDE$ are congruent employing $SAS$ congruence. Now letting angle $ACD$ = angle $BCD=y$ , triangles $ACE$ and $BDE$ being congruent means $CE=DE$ therefore angle $CDE$ = angle $DCE=y$ . Moreover, angle $BDE$ = angle $ACE$ = angle $ACD$ + angle $BCD=y+y=(1+1)y=2y$ .

With angle $BAC$ being twice of angle $ABC$ , angle $BAC=2x$ therefore we have:

$ACD+DAC=BDC$

$ACD+DAC=BDE+CDE$

$y+2x=2y+y$

$2x=2y$

$x=y$

Considering triangle $ABC$ , we also have:

$BAC+ABC+ACB=\pi$

$2x+x+2y=\pi$

$(2+1)x+2y=\pi$

$3x+2y=\pi$

Therefore with $x=y$ ,

$3x+2x=\pi$

$(3+2)x=\pi$

$5x=\pi$

$x=\frac{\pi}{5}$

$=>N=\boxed{5}$