parmenides51 wrote: The circle of diameter $LEB$ intersects the segment $BC$ at $O$. Prove that the lines $HD$ , $LE$ and $BC$ are concurrent if and only if $EO = EC$. A strange diameter $L\ E\ B$ ?

wording has been corrected, thanks for the correction

Given $ABCD\ :\ A(0,0),B(a,b),C(a+d,b),D(d,0)$. Point $E(\frac{a+d}{2},\frac{b}{2})$. Midpoint of $ED\ :\ (\frac{a+3d}{4},\frac{b}{4})$. Equation of the circle: $x^{2}+y^{2}-\frac{a+3d}{2}x-\frac{b}{2}y+\frac{d(a+d)}{2}=0$. This circle intersects the x-axis in the point $L(\frac{a+d}{2},0)$ and the line $AC$ in the point $H(\frac{d(a+d)^{2}}{(a+d)^{2}+b^{2}},\frac{bd(a+d)}{(a+d)^{2}+b^{2}})$. The circumcircle of $L,E,B$ intersects the line $AB$ a second time in the point $O(\frac{a^{2}+2b^{2}-d^{2}}{2(a-d)},b)$. The lines $BC\ :\ y=b\ ,\ LE\ :\ x=\frac{a+d}{2}$ and $HD\ :\ y=\frac{a+d}{b}(x-d)$ are concurrent if $2b^{2}=d^{2}-a^{2}$. In this case, $EO=EC$.