Degree of an aforementioned angle.

Let us choose $\overline {AC}$ to be along $X$ -axis, $\overline {AB}$ along $Y$ -axis, $|\overline {AB}|=c, |\overline {AC}|=b$ . Then the coordinates of $A, B, C, D$ and $E$ are $(0,0), (0,c), (b, 0), (-c, -b)$ and $(b+c,0)$ respectively. Slope of $\overline {BD}$ is $\dfrac{b+c}{c}$ . So, that of $\overline {AK}$ is $-\dfrac{c}{b+c}$ , and it's equation is $y=-\dfrac{c}{b+c}x$ . Equation of $\overline {DE}$ is $y=\dfrac{b}{b+2c}(x-b-c)$ . Solving these two equations we get the coordinates of $K$ as $(\dfrac{b(b+c)^2}{b^2+2bc+2c^2}, -\dfrac{bc(b+c)}{b^2+2bc+2c^2})$ . Therefore, slope of $\overline {CK}$ is $\dfrac{b+c}{c}$ and that of $\overline {EK}$ is $\dfrac{b}{b+2c}$ . Hence, $\angle {CKE}=\tan^{-1} (\dfrac{\dfrac{b+c}{c}-\dfrac{b}{b+2c}}{1+\dfrac{b(b+c)}{c(b+2c)}})=\boxed {45\degree}$ . (We observe that $\overline {BD}$ and $\overline {CK}$ are parallel)