Let $A(0,0),B(2b,0),C(2b,2b),D(0,2b)$, then $N(0,b)$. $\triangle ABE\ :\ E(b,-b)$. The line $CE\ :\ y-2b=3(x-2b)$ cuts $AB$ in the point $M(\frac{4b}{3},0)$. The line $CN\ :\ y-2b=\frac{1}{2}(x-2b)$ cuts $AB$ in the point $P(-2b,0)$. The line $PE\ :\ y=-\frac{1}{3}(x+2b)$ cuts the line $MN\ :\ y-b:-\frac{3}{4}x$ in the point $F(4b,-2b)$. Choose $Q(\lambda,-\frac{1}{2}(\lambda+2b)$ on the line $PE$. Because the slope of the line $CE\ :\ m_{CE}=3=\tan \widehat{CMB}$, is $\tan \widehat{MCB}=\tan(\frac{\pi}{2}-\widehat{CMB})=\cot \widehat{CMB}=\frac{1}{3}$. The slope of the line $QC\ :\ m_{QC}=\frac{8b+\lambda}{3(2b-\lambda)}$ and $\tan \widehat{QCM}=\frac{3-\frac{8b+\lambda}{3(2b-\lambda}}{1+\frac{8b+\lambda}{2b-\lambda}}=\frac{b-\lambda}{3b}$. Equal angles $\Rightarrow frac{b-\lambda}{3b}=\frac{1}{3} \Rightarrow \lambda=0$. So, point $Q(0,-\frac{2b}{3})$. Slope of the line $CF\ :\ m_{CF}=-2$, slope of the line $MQ\ :\ m_{MQ}=\frac{1}{2}$, hence $MQ \bot CF$.