Given the quadrilateral $ABCD\ :\ A(0,0),B(b,0), AC=c,AD=d, \angle BAC=\alpha$, then $C(c\cos \alpha,c\sin a)$ and $D(-d\sin \alpha,d\cos a)$. Slope of the line $BC\ :\ m_{BC}=\frac{c\sin \alpha}{c\cos \alpha-b}$, then the slope of $CE\ :\ m_{CE}=\frac{b-c\cos \alpha}{c\sin \alpha}$. $CE$ is the bisector of the lines $AC\ :\ \sin \alpha \cdot x-\cos \alpha \cdot y$ and $CD\ :\ y-c\sin \alpha=\frac{c\sin \alpha-d\cos \alpha}{c\cos \alpha+d\sin \alpha}(x-c\cos \alpha)$, the equation of this bisector $[(c+w)\sin \alpha-d\cos \alpha]x-[d\sin \alpha+(c+w)\cos \alpha]y+cd-c^{2}=0$ with $w=\sqrt{c^{2}+d^{2}}$. The slope of this bisector equals $\frac{b-c\cos \alpha}{c\sin \alpha}$, giving $b=\frac{c(c+w)}{d\sin \alpha+(c+w)\cos a}$. The lines $BD$ and $CE$ cut in the point $E(\frac{c(d\sin \alpha+b)-b\cos \alpha(2d\sin \alpha+b)}{b(c\cos \alpha-d\sin \alpha-b)},-\frac{d\cos \alpha}{d\sin \alpha+b}(x_{E}-b))$. Calculating the slope of the line $AE$, using the expression of $b$, we find $m_{AE}=\tan 2\alpha$, so $AC$ is the bisector of $\angle BAE$.