Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$. Choose $D(\lambda,0)$. Points $E(\frac{b\mu}{a-b},\mu)$ and $F(c-\frac{c\mu}{a},\mu)$ on the line $EF\ :\ y=\mu$. $EF=(b+c)(1-\frac{\mu}{a})$ and $BC=b+c \Rightarrow \frac{EF}{BC}=1-\frac{\mu}{a}$. Point $G(\frac{\mu \lambda}{a}-b+\frac{b\mu}{a},0)$. $AD=\sqrt{\lambda^{2}+a^{2}}$ and $EG=\frac{\mu \sqrt{\lambda^{2}+a^{2}}}{a} \Rightarrow \frac{EG}{AD}=\frac{\mu}{a}$. Midpoints $M(\frac{\lambda}{2},\frac{a}{2})$ and $N(\frac{c-b}{2},0)$. Midpoint of $FG\ :\ P(\frac{c-\frac{c\mu}{a}+\frac{\mu \lambda}{a}-b+\frac{b\mu}{a}}{2},\frac{\mu}{2})$ lies on the line $MN\ :\ y=\frac{a}{\lambda-c+b}(x-\frac{c-b}{2})$.