Let $A(0,a),B(-b,0),C(c,0)$; first condition $\Rightarrow D(-c,0)$. $A_{1}(0,0)$. The line $BH\ :\ y=\frac{c}{a}(x+b)$ cuts the y-axis in the point $H(0,\frac{bc}{a})$. The line $BH$ cuts the line $AC\ :\ y=-\frac{a}{c}(x-c)$ in the point $B_{1}(\frac{c(a^{2}-bc)}{a^{2}+c^{2}},\frac{ac(b+c)}{a^{2}+c^{2}})$. The line $A_{1}B_{1}\ :\ y=\frac{a(b+c)}{a^{2}-bc} \cdot x$ cuts the line $DH\ :\ y=\frac{b}{a}(x+c)$ in the point $E(\frac{b(a^{2}-bc)}{a^{2}+b^{2}},\frac{ab(b+c)}{a^{2}+b^{2}})$. The circle, through $B,D,B_{1}$ has midpoint $O(-\frac{b+c}{2},a)$ and equation $x^{2}+(b+c)x+y^{2}-2ay+bc=0$. The coordinates of $E$ satisfy this equation.