Given the circle, midpoint $O(0,0)$ and radius $r$. Points $B$ and $C$ on the line $y=-b, \quad b>0$. Points $A(0,r)$ and $E(0,-r)$; choose $D(r\cos \alpha,r\sin \alpha)$. The line $AD\ :\ y-r=\frac{\sin \alpha-1}{\cos \alpha} \cdot x$ cuts $BC$ in the point $S(\frac{(b+r)\cos \alpha}{1-\sin \alpha},-b)$. The line $DE\ :\ y+r=\frac{\sin \alpha+1}{\cos \alpha} \cdot x$ cuts $BC$ in the point $T(\frac{(b-r)\cos \alpha}{1+\sin \alpha},-b)$. Midpoint $F(\frac{b\sin \alpha+r}{\cos \alpha},-b)$. Equation of the circumcircle of $\triangle ODF\ :\ x^{2}+y^{2}-\frac{b\sin \alpha+r}{\cos \alpha} \cdot x+by=0$. On this circle, the points $F$ and $I(0,-b)$. Equation of the line $t // OD$, through $I\ :\ y+b=\tan \alpha \cdot x$. $t$ cuts in $M(\ (b+r)\cos \alpha,(b+r)\sin \alpha-b)$ and $N(\ (r-b)\cos \alpha,(r-b)\sin \alpha-b)$. The area of $\triangle DMN$ equals $b^{2}\cos \alpha$ and has a maximum $b^{2}$ if $\alpha=0$. The perpendicular from $D$ to $ST\ :\ x=r\cos \alpha$ cuts $MN$ in his midpoint $(r\cos \alpha,r\sin \alpha-b)$.