Given $A(0,d),B(0,0),C(c,0),D(c,d)$; choose $K(\lambda,0)$. On the line $CD$, point $N(c,n)$, then $\tan \beta=\tan \angle NKC=\frac{n}{c-\lambda}$. Given $\angle NKL=\alpha$, a fixed angle. $\tan(\alpha+\beta)=\frac{\tan \alpha+\frac{n}{c-\lambda}}{1-\tan \alpha \cdot \frac{n}{c-\lambda}}=\frac{(c-\lambda)\tan \alpha+n}{c-\lambda-n\tan \alpha}$, the slope of the line $KL$. The line $KL\ y=\frac{(c-\lambda)\tan \alpha+n}{c-\lambda-n\tan \alpha} \cdot (x-\lambda)$ cuts $AB$ in the point $L(0,-\frac{\lambda[(c-\lambda)\tan \alpha+n]}{c-\lambda-n\tan \alpha})$. Condition $KL=KN \Rightarrow n=\frac{\lambda+(c-\lambda)\cos \alpha}{\sin \alpha}$. Then $L(0,\frac{c-\lambda+\lambda \cos \alpha}{\sin \alpha})$. The midpoint of $LN$ is the fixed point $(\frac{c}{2},\frac{c}{2}\cot \frac{\alpha}{2})$.