We'll use Cartesian Coordinates to give a proof for this claim! Set $EH_1$ as the $Y- $axis and $DF$ as the $X-$axis Set $D \equiv (c,0)$ and $F \equiv (b,0)$ and WLOG, we can safely assume $E \equiv (0,1)$, because just magnification of the triangle won't change the configuration and won't affect our calculations and Let $b,c \in R$ Since, $H_1$ is the origin, hence, $H_1 \equiv (0,0)$ and really easy to see that the equation of the perpendicular bisector of $DF$ is $x=\frac{b+c}{2}$ $I$ lies on the perpendicular bisector of $DF$: Proof uses the fact that, $BFID$ is cyclic $\Longrightarrow$ $\angle IFD=\angle IDF=\frac{1}{2}\angle ABC$, Using this fact, $I \equiv \left( \frac{b+c}{2} ,y \right) $, we need to calculate the y coordinate of $I$ $\textbf{\underline{Coordinate for I}}$: We know $I\equiv \left( \frac{b+c}{2} , y \right)$, Since $I$ is the circumcenter of $\omega$, therefore, $EI=DI$, hence, by distance formula we get the following: $$EI=DI$$$$\sqrt{\left( \frac{b+c}{2} +0 \right) ^2 +(y-1)^2}=\sqrt{ \left( \frac{b+c}{2} -c \right) ^2 +y^2 }$$And after simplifying this gives $\Longrightarrow$ $y=\frac{bc+1}{2}$, Hence, the coordinate for $I$ is $$I \equiv \left( \frac{b+c}{2} ,\frac{bc+1}{2} \right) $$ We'll use this notation to represent the equation of line throughout the solution: For a line $XY$ with equation $ax+by+c=0$ will be represented as $$XY : ax+by+c=0$$ \end{flushleft} \begin{flushleft} We'll use the famous formula from coordinate geometry to find out the equation of line when the coordinates for two points on it is given, The formula is as follows: \par \end{flushleft} \par \begin{flushleft} \textbf{\underline{Formula 1}}: If $X \equiv (x_1,y_1)$ and $Y \equiv (x_2,y_2)$, then, $$XY: \frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}$$Which can be re-written as: $$XY: \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$$ \end{flushleft} \par \textbf{\underline{Equation for line IE}}: We know that the coordinate for $I \equiv \left( \frac{b+c}{2} ,\frac{bc+1}{2} \right)$ and $E \equiv (0,1)$, Hence, $$IE: \frac{y-1}{\frac{bc+1}{2}-1}=\frac{x-0}{\frac{b+c}{2}-0}$$$$IE: y-1=x\left( \frac{bc-1}{b+c} \right)$$ $$IE: y=1+\left( \frac{bc-1}{b+c} \right) x $$ \par \textbf{\underline{Equation for line IF}}: We know that the coordinate for $I \equiv \left(\frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $F \equiv (b,0)$, Hence, $$IF : \frac{y-0}{\frac{bc+1}{2}-0}=\frac{x-b}{\frac{b+c}{2}-b}$$$$IF : y=\left(\frac{x-b}{\frac{c-b}{2}} \right) \left( \frac{bc+1}{2} \right)$$$$IF : y=\left( \frac{bc+1}{c-b} \right) x +\frac{b^2c+b}{b-c}$$ \textbf{\underline{Equation for line ID}}: We know that the coordinate for $I \equiv \left( \frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $D \equiv (c,0)$, Hence, $$ID : \frac{y-0}{\frac{bc+1}{2}-0}=\frac{x-c}{\frac{b+c}{2}-c}$$$$ID : y=\left( \frac{x-c}{\frac{b-c}{2}} \right) \left( \frac{bc+1}{2} \right)$$$$ID : y=x \left( \frac{bc+1}{b-c} \right) -\frac{c(bc+1)}{b-c} $$ \par \begin{flushleft} Note that, $AC \perp EI$, $AB \perp IF$ and $BC \perp ID$ \end{flushleft} \par \begin{flushleft} We'll use the famous formula from coordinate geometry to find out the slope of line when the coordinates of two points lying on the line are given, The formula is as follows: \par \end{flushleft} \par \begin{flushleft} \textbf{\underline{Formula 2}}: If $X \equiv (x_1,y_1)$ and $Y \equiv (x_2,y_2)$ and Let $m_{XY}$ represent the slope of the line $XY$, then, $$m_{XY}=\frac{y_1-y_2}{x_1-x_2}$$\end{flushleft} \textbf{\underline{Slope for IF}}: The coordinate for $I \equiv \left(\frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $F \equiv (b,0)$ $$m_{IF}=\frac{\frac{bc+1}{2}-0}{\frac{b+c}{2}-b}$$$$m_{IF}=\frac{bc+1}{c-b}$$ \textbf{\underline{Slope for IE}}: The coordinate for $I \equiv \left( \frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $E \equiv (0,1)$ $$m_{IE}=\frac{\frac{bc+1}{2} -1}{\frac{b+c}{2}-0}$$$$m_{IE}=\frac{bc-1}{b+c}$$ \textbf{\underline{Slope for ID}}: The coordinate for $I \equiv \left( \frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $D \equiv (c,0)$ $$m_{ID}=\frac{\frac{bc+1}{2}-0}{\frac{b+c}{2}-c}$$$$m_{ID}=\frac{bc+1}{b-c}$$ \begin{flushleft} We'll state another famous result from coordinate geometry \par \end{flushleft} \par \begin{flushleft} \textbf{\underline{Formula 3}}: If $X_1Y_1$ and $X_2Y_2$ are perpendicular to each other, then, $$m_{X_1Y_1} \cdot m_{X_2Y_2}=-1$$\par \end{flushleft} \par \textbf{\underline{Slope for AB}}: Applying Formula 3 $$m_{AB}=\frac{b-c}{bc+1}$$ \textbf{\underline{Slope for BC}}: Applying Formula 3 $$m_{BC}=\frac{c-b}{bc+1}$$ \textbf{\underline{Slope for AC}}: Applying Formula 3 $$m_{AC}=\frac{-(b+c)}{bc-1}$$ \begin{flushleft} We'll state another well known formula from coordinate geometry to find out the equation of line when the coordinates for a point lying on the line and the slope of line is given \par \end{flushleft} \par \begin{flushleft} \textbf{\underline{Formula 4}}: If $X_1 \equiv (x_1,y_1)$ lies on $\ell$ and $m_{\ell}$ is the slope of $\ell$, then, $$\ell : y-y_1=m_{\ell} (x-x_1)$$\end{flushleft} \textbf{\underline{Equation for line AB}}: $F$ lies on $AB$ and $F \equiv (b,0)$ Hence, $$AB: y-0=\left( \frac{b-c}{bc+1} \right) (x-b)$$$$AB: y=\left( \frac{b-c}{bc+1} \right) (x-b)$$ \textbf{\underline{Equation for line BC}}: $D$ lies on $BC$ and $D \equiv (c,0)$ $$BC: y-0=\left( \frac{c-b}{bc+1} \right) (x-c)$$$$BC: y=\left( \frac{c-b}{bc+1} \right) (x-c)$$ \textbf{\underline{Equation for line AC}}: $E$ lies on $AC$ and $E \equiv (0,1)$ $$AC: y-1=\frac{-(b+c)}{bc-1} (x-0)$$$$AC: y=1-x\left( \frac{b+c}{bc-1} \right)$$ \textbf{\underline{Coordinate for X}}: now recall that $H_1$ is the feet of perpendicular from $E$ to $DF$ and $EH_1 \cap BC=X$ and also, $EH_1$ is the y-axis, and hence, x- coordinate for $X$ is $0$ $\Longrightarrow$ $X \equiv (0,y)$ and since, $X \in BC$, we have, $$BC:y=\left( \frac{c-b}{bc+1} \right) (x-c)$$$$y=-c \left( \frac{c-b}{bc+1} \right)$$$$y=\frac{bc-c^2}{bc+1}$$$$\Longrightarrow X \equiv \left( 0, \frac{bc-c^2}{bc+1} \right)$$ \textbf{\underline{Equation for line DE}}: Recall, $D \equiv (c,0)$ and $E \equiv (0,1)$, Hence, using Formula 1, $$DE:\frac{y-0}{1-0}=\frac{x-c}{0-c}$$$$DE: y=1-\frac{x}{c}$$ \textbf{\underline{Equation for line EF}}: Recall, $E \equiv (0,1)$ and $F \equiv (b,0)$, Hence, using Formula 1, $$EF: \frac{y-0}{1-0}=\frac{x-b}{0-b}$$$$EF: y=1-\frac{x}{b}$$ \textbf{\underline{Slope for DE}}: $D \equiv (c,0)$ and $E \equiv (0,1)$, $$m_{DE}=\frac{0-1}{c-0}$$$$m_{DE}=\frac{-1}{c}$$ \textbf{\underline{Equation for line $FH_2$}}: $F \equiv (b,0)$ Notice, $FH_2 \perp ED$, hence, $$m_{FH_2} \cdot m_{DE}=-1$$$$m_{FH_2}=c$$And since, $F$ lies on $FH_2$, therefore, $$FH_2 : y-0=c(x-b)$$$$FH_2 : y=c(x-b)$$ \textbf{\underline{Coordinate for Y}}: Recall $FH_2 \cap BC=Y$, we have, $FH_2 : y=c(x-b)$ and $BC: y=\left( \frac{c-b}{bc+1} \right) (x-c)$, Hence, solving equations: should yield coordinates for $Y$ $$FH_2 : y=c(x-b) -----(1)$$$$BC: y=\left( \frac{c-b}{bc+1} \right) (x-c) ------(2)$$ Equating $(1)$ and $(2)$, we have, $$c(x-b)=\frac{(c-b)(x-c)}{bc+1}$$$$\frac{1}{bc+1}=\frac{c(x-b)}{(c-b)(x-c)}=\frac{cx-bc}{cx+bc-bx-c^2}$$$$cx+bc-bx-c^2=(cx-bc)(bc+1)$$$$cx+bc-bx-c^2=bc^2x+cx-b^2c^2-bc$$$$2bc-c^2-bx=bc^2x-b^2c^2$$$$2bc-c^2+b^2c^2=bc^2x+bx$$$$x=\frac{2bc-c^2+b^2c^2}{b(c^2+1)}$$ and, $$y=c(x-b)$$$$y=c \left( \frac{2bc-c^2+b^2c^2}{b(c^2+1)} -b \right)$$$$y=c \left( \frac{2bc-c^2+b^2c^2-b^2c^2-b^2}{bc^2+b} \right)$$$$y=\frac{-c(b-c)^2}{bc^2+b}$$ Hence, $$Y \equiv \left( \frac{2bc+b^2c^2-c^2}{b(c^2+1)} , \frac{-c(b-c)^2}{b(c^2+1)} \right)$$ \par \begin{flushleft} Let $D^*$ be a point diametrically opposite to $D$ on $\omega$, or inshort, Let $D^*$ be the $D-$antipode in $\omega$, hence, $\angle D^*FD=\angle EH_1D=90^{\circ}$ $\Longrightarrow$ $FD^*||EH_1$ $$\Longrightarrow D^* \equiv (b,y)$$and since, $D^* \in ID$ and $ID:y=x \left( \frac{bc+1}{b-c} \right) -\frac{c(bc+1)}{b-c} $ $$y=x \left( \frac{bc+1}{b-c} \right) -\frac{c(bc+1)}{b-c} $$$$y=(x-c) \left( \frac{bc+1}{b-c} \right) $$$$y=(b-c) \left( \frac{bc+1}{b-c} \right)$$$$y=bc+1$$Hence, $$D^* \equiv (b,bc+1)$$\par \end{flushleft} \par \textbf{\underline{Coordinate for A}}:Recall, $AB \cap AC=A$ $$AB: y=\left( \frac{b-c}{bc+1} \right) (x-b)$$$$AC:y=1-x\left( \frac{b+c}{bc-1} \right)$$ $$\left( \frac{b-c}{bc+1} \right) (x-b)=1-x\left( \frac{b+c}{bc-1} \right)$$$$\frac{bc-1}{bc+1}=\frac{(bc-1)-x(b+c)}{(b-c)(x-b)}$$$$(b-c)(x-b)(bc-1)=(bc+1)(bc-1)-x(b+c)(bc+1)$$$$(b^2c-bc^2-b+c)(x-b)=b^2c^2-1-(b^2c+b+bc^2+c)x$$$$b^2c^2-1-(b^2c+b+bc^2+c)x=x(b^2c-bc^2-b+c) -b(b^2c-bc^2-b+c)$$$$b^2c^2-1+b(b^2c-bc^2-b+c)=x(b^2c-bc^2-b+c+b^2c+b+c+bc^2)$$$$b^2c^2-1+b^3c-b^2c^2-b^2+bc=x(2b^2c+2c)$$$$b^3c-b^2+bc-1=x(2b^2c+2c)$$$$(bc-1)(b^2+1)=x(2c)(b^2+1)$$$$x=\frac{bc-1}{2c}$$And, $$y=1-x\left( \frac{b+c}{bc-1} \right)$$$$y=1-\frac{b+c}{2c}$$$$y=\frac{c-b}{2c}$$Hence, $$A \equiv \left( \frac{bc-1}{2c} , \frac{c-b}{2c} \right) $$ \textbf{\underline{Equation for line AD}}: Recall, $D \equiv (c,0)$ $$AD: \frac{y-0}{\frac{c-b}{2c}-0}=\frac{x-c}{\frac{bc-1}{2c}-c}$$$$AD: y=\frac{x-c}{\frac{bc-1-2c^2}{2c}} \left(\frac{c-b}{2c} \right)$$$$AD: y =\frac{(x-c)(c-b)}{bc-1-2c^2}$$\par \begin{flushleft} Recall $AZ \cap EF=K$, but since, $Z \in AD$, hence, $AD \cap EF=K$ \par \end{flushleft} \par \textbf{\underline{Coordinate for K}}: Recall, $AD \cap EF=K$, We have, $$AD: y =\frac{(x-c)(c-b)}{bc-1-2c^2}$$$$EF: y=1-\frac{x}{b}$$ Hence, $$1-\frac{x}{b}=\frac{(x-c)(c-b)}{bc-1-2c^2}$$ $$1-\frac{x}{b}=\frac{x(c-b)}{bc-1-2c^2} +\frac{c(b-c)}{bc-1-2c^2}$$$$1-\frac{(bc-c^2)}{bc-1-12c^2}=\frac{x}{b}+\frac{x(c-b)}{bc-1-2c^2}$$$$\frac{bc-1-2c^2-bc+c^2}{bc-1-2c^2}=\frac{x(bc-1-2c^2+bc-b^2)}{b(bc-1-2c^2)}$$$$\frac{-(c^2+1)}{bc-1-2c^2}=\frac{x(2bc-1-2c^2-b^2)}{b(bc-1-2c^2}$$$$x=\frac{-b(c^2+1)}{2bc-1-2c^2-b^2}$$$$x=\frac{b(c^2+1)}{2c^2+b^2+1-2bc}$$And, $$y=1-\frac{x}{b}$$$$y=1-\frac{b(c^2+1)}{b(2c^2+b^2+1-2bc)}$$$$y=\frac{2c^2+b^2+1-2bc-c^2-1}{2c^2+b^2+1-2bc}$$$$y=\frac{(b-c)^2}{2c^2+b^2+1-2bc}$$Hence, $$K \equiv \left( \frac{b(c^2+1)}{2c^2+b^2+1-2bc} , \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)$$\par \textbf{\underline{Coodinate for Z}}: Note, that, since, $DD^*$ is the diameter of $\omega$, hence, $ZD^* \perp ZD$ or $ZD^* \perp AD$, therefore, $$m_{ZD^*} \cdot m_{AD}=-1$$Now we have eariler found out that, $$AD: y=\frac{(x-c)(c-b)}{bc-1-2c^2}$$$$AD:y=\frac{x(c-b)}{bc-1-2c^2} -\frac{c(b-c)}{bc-1-2c^2}$$We know that equation of a line can be represented in the form of $y=mx+b$, where $m$ is the slope of the line and $b$ is the $y-$intercept, hence using this we get, $$m_{AD}=\frac{(c-b)}{bc-1-2c^2}$$ Hence, $$ m_{ZD^*}=\frac{-1}{m_{AD}}=\frac{-1}{\frac{c-b}{bc-1-2c^2}}$$ $$m_{ZD^*}=\frac{2c^2+1-bc}{c-b}$$Let $Z \equiv (x',y')$ and recall that $D^* \equiv (b,bc+1)$, hence ,we can find equation of $ZD^*$, $$ZD^*: \frac{y-y'}{bc+1-y'}=\frac{x-x'}{b-x'}$$$$ZD^*:y=y'+\left( \frac{x-x'}{b-x'} \right) (bc+1-y')$$Using the earier we stated that, any line can be written in the form of $y=mx+b$, so using this, $$m_{ZD^*}=\frac{bc+1-y'}{b-x'}$$but, we eariler found out that, $$m_{ZD^*}=\frac{2c^2+1-bc}{c-b}$$$$\Longrightarrow \frac{2c^2+1-bc}{c-b}=\frac{bc+1-y'}{b-x'} ----(3)$$And, Since, $Z \in AD$, $$AD: y=\frac{(x-c)(b-c)}{bc-1-2c^2}$$$$y'=\frac{(x-c)(b-c)}{bc-1-2c^2}----(4)$$Solving $(3)$ and $(4)$ should yield the coordinates for $Z$, $$(3) \longrightarrow$$$$\frac{2c^2+1-bc}{c-b}=\frac{bc+1-y'}{b-x'}$$$$(2c^2+1-bc)(b-x')=(bc+1)(c-b)-y'(c-b)$$$$y'=\frac{(2c^2+1-bc)(b-x')+(bc+1)(b-c)}{b-c} -----(5)$$ Therefore, dividing $(4)$ and $(5)$, we have, $$\frac{y'}{y'}=\frac{\frac{(x'-c)(c-b)}{bc-1-2c^2}}{\frac{(2c^2+1-bc)(b-x')+(bc+1)(b-c)}{b-c}}$$$$\frac{bc-1-2c^2}{b-c}=\frac{(x'-c)(c-b)}{(2c^2+1-bc)(b-x')+(bc+1)(b-c)}$$$$\frac{bc-1-2c^2}{b-c}=\frac{x'(c-b) +bc-c^2}{2bc^2+b-b^2c-x'(2c^2+1-bc)+b^2c+b-bc^2-c}$$$$\frac{bc-1-2c^2}{b-c}=\frac{x'(c-b) +bc-c^2}{bc^2+2b-c-x'(2c^2+1-bc)}$$$$(bc-1-2c^2)(bc^2+2b-c-x'(2c^2+1-bc))=x'(c-b)(b-c)+(bc-c^2)(b-c)$$$$(bc-1-2c^2)(bc^2+2b-c)+x'(bc-1-2c^2)^2=c(b-c)^2-x'(b-c)^2$$$$b^2c^3+2b^2c-bc^2-bc^2-2b+c-2bc^4-4bc^2+2c^3+x'(bc-1-2c^2)^2=b^2c-2bc^2+c^3-x'(b-c)^2$$$$b^2c^3+2b^2c-bc^2-bc^2-2b+c-2bc^4-4bc^2+2c^3-b^2c+2bc^2-c^3=-x'((b-c)^2+(bc-1-2c^2))$$$$(b^2c-2bc^2-2b+c)(c^2+1)=-x'(b^2-2bc+c^2+b^2c^2+1+4c^4-2bc+4c^2-4bc^3)$$$$(b^2c-2bc^2-2b+c)(c^2+1)=-x'((c^2+1)(b^2-4bc+4c^2+1))$$$$x' \equiv - \left( \frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} \right)$$ And, Since, as above stated, $$y'=\frac{(x'-c)(c-b)}{bc-1-2c^2}$$$$y'=\frac{(-(\frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1})-c)(c-b)}{bc-1-2c^2}$$$$y'=\frac{-(\frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} +c)(c-b)}{bc-1-2c^2}$$$$y'=\frac{(\frac{b^2c-2bc^2-2b+c+b^2c-4bc^2+4c^3+c}{b^2-4bc+4c^2+1})(b-c)}{bc-1-2c^2}$$$$y'=\frac{2(b-c)(bc-1-2c^2)(b-c)}{(bc-1-2c^2)(b^2-4bc+4c^2+1)}$$$$y' \equiv \frac{2(b-c)^2}{b^2-4bc+4c^2+1}$$Hence, Combining we have the coordinates for $Z$ $$Z \equiv \left( - \left( \frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} \right) , \left( \frac{2(b-c)^2}{b^2-4bc+4c^2+1} \right) \right)$$ \par \textbf{\underline{Equation for line $ZD^*$ }}: Recall, $D^* \equiv (b,bc+1)$ and we have found out the coordinate for $Z$ above, Hence, $$ZD^* : \frac{y-(bc+1)}{\frac{2(b-c)^2}{b^2-4bc+4c^2+1} -(bc+1)}=\frac{x-b}{- \left( \frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} \right) -b}$$$$ZD^*: \frac{y-(bc+1)}{2(b-c)^2-(bc+1)(b^2-4bc+4c^2+1)}=\frac{x-b}{-(b^2c-2bc^2-2b+c+b^3-4b^2c+4bc^2+b)}$$$$ZD^*: \frac{y-(bc+1)}{-(b^2-2bc-1)(bc-2c^2-1)}=\frac{x-b}{-(b-c)(b^2-2bc-1)}$$ $$ZD^*: y=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$ \par \textbf{\underline{Equation for line AX}}: Recall, $A \equiv \left( \frac{bc-1}{2c} , \frac{c-b}{2c} \right)$ and $X \equiv \left( 0, \frac{bc-c^2}{bc+1} \right) $, Hence, $$AX: \frac{y- \frac{c-b}{2c}}{\frac{bc-c^2}{bc+1}-(\frac{c-b}{2c})}=\frac{x-\frac{bc-1}{2c}}{0-(\frac{bc-1}{2c})}$$$$AX: \frac{y-\frac{c-b}{2c}}{\frac{(bc-c^2)(2c)-(c-b)(bc+1)}{(bc+1)(2c)}} =\frac{x-(\frac{bc-1}{2c})}{-(\frac{bc-1}{2c})}$$$$AX: \frac{y-(\frac{c-b}{2c})}{\frac{b^2c+bc^2+b-2c^3-c}{bc+1}}=\frac{x-(\frac{bc-1}{2c})}{1-bc}$$$$AX: y-\left(\frac{c-b}{2c} \right)=\left( \frac{2cx-(bc-1)}{2c(1-bc)} \right) \left( \frac{b^2c+bc^2+b-2c^3-c)}{bc+1} \right)$$ $$AX: y=\frac{c-b}{2c}- \left( \frac{(2cx-(bc-1))(b^2c+bc^2+b-2c^3-c)}{2c(b^2c^2-1)} \right)$$ \par \textbf{\underline{Coordinate for L}}: Recall, that, $AX \cap ZD^*=L$ We have already found the equation for line $AX$ and line $ZD^*$, so now, we can find out the coordinates for $L$, $$AX: y=\frac{c-b}{2c}- \left( \frac{(2cx-(bc-1))(b^2c+bc^2+b-2c^3-c)}{2c(b^2c^2-1)} \right)$$$$ZD^*: y=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$Equating $y$, we get, $$\Longrightarrow \frac{c-b}{2c}- \left( \frac{(2cx-(bc-1))(b^2c+bc^2+b-2c^3-c)}{2c(b^2c^2-1)} \right)=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$$$\Longrightarrow -(b-c)^2(b^2c^2-1)-(2cx+1-bc)(b^2c+bc^2+b-2c^3-c)(b-c)$$$$=2c(bc+1)(b-c)(b^2c^2-1)+(2c)(bc-1-2c^2)(x-b)(b^2c^2-1)$$ $$\Longrightarrow -2b^3c^4x+4b^2c^5x-6c^3x-2cx-2b^3c^2x-2b^2cx+6bc^4x+6bc^2x-4c^5x+2b^2c^3x$$$$=2b^3c^3+2c^2+2b^3c^5-4bc+2b^2c^4+2b^2c^2-2bc^5-6bc^3+2c^4$$ $$\Longrightarrow -2c(c^2+1)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)x=2c(c^2+1)(bc-1)(b^2c+2b-c)$$$$\Longrightarrow x=\frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}$$ Now, we can use the value of $x$ to find out the value of $y$, $$y=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$$$y=\frac{(bc+1)(b-c) + (bc-2c^2-1)(x-b)}{b-c}$$$$y=\frac{(b^2-bc^2+b-c)+(bc-2c^2-1) \left(\frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}-b \right)}{b-c}$$$$y=\frac{(bc+1)(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)-(bc-2c^2-1)(b^4c-b^3c^2+b^3-2b^2c+bc^2-b+c)}{(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}$$$$y=\frac{2b^4c-4b^3c^2+2b^3+2b^2c^3-4b^2c+2bc^2}{(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}$$$$y=\frac{2b(b-c)^2(bc+1)}{(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}$$$$y=\frac{2b(b-c)(bc+1)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}$$ \par \par $$L \equiv \left( \frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} , \frac{2b(b-c)(bc+1)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} \right)$$ \par \begin{flushleft} Now we'll calculate the equation for $YK$ \end{flushleft} \par \textbf{\underline{Equation for line YK}}: Recall, $Y \equiv \left( \frac{2bc+b^2c^2-c^2}{b(c^2+1)} , \frac{-c(b-c)^2}{b(c^2+1)} \right)$ and \par $K \equiv \left( \frac{b(c^2+1)}{2c^2+b^2+1-2bc} , \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)$ Hence, we have, $$YK : \frac{y-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{-c(b-c)^2}{b(c^2+1)}-\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}=\frac{x-\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{2bc+b^2c^2-c^2}{b(c^2+1)}-\left(\frac{b(c^2+1)}{2c^2+b^2+1-2bc}\right)} $$$$YK : \frac{y-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{c(b-c)^2}{b(c^2+1)}+\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}=\frac{x-\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{b(c^2+1)}{2c^2+b^2+1-2bc}-\left(\frac{2bc+b^2c^2-c^2}{b(c^2+1)} \right)} $$ Now if the coordinates of $L$ satisfy equation of $YK$, then we have indeed proved that $L$ lies on $YK$, so we'll do it by first computing the RHS and then LHS and then if RHS equals LHS, we're done with our claim, So here we go!, \par \textbf{\underline{LHS}}:\par $$\Longrightarrow \frac{y-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{c(b-c)^2}{b(c^2+1)}+\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}$$$$\Longrightarrow \frac{\left( \frac{2b(b-c)(bc+1)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} \right)-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{c(b-c)^2}{b(c^2+1)}+\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}$$$$\Longrightarrow \frac{\frac{(b^2-2bc+2c^2+1)(2b^2-2bc)(bc+1)-(b-c)^2(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}{(b^3c+b^2-2b^2c^2-3bc+2c^2+1)(b^2-2bc+2c^2+1)}}{\frac{c(b-c)^2(b^2-2bc+2c^2+1)+(b-c)^2(bc^2+b)}{(bc^2+b)(b^2-2bc+2c^2+1)}}$$$$\Longrightarrow \frac{\frac{b^4c-b^3c^2+b^3+2b^2c^3+2b^2c-bc^2+b+2c^3+c}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}}{\frac{(b-c)(b^2c-bc^2+b+2c^3+c)}{bc^2+b}}$$$$\Longrightarrow \frac{\frac{b^4c-b^3c^2+b^3+2b^2c^3+2b^2c-bc^2+b+2c^3+c}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}}{\frac{b^3c-2b^2c^2+b^2+3bc^3-2c^4-c^2}{bc^2+b}} ------(6)$$ \par \textbf{\underline{RHS}}: \par $$\Longrightarrow \frac{x-\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{b(c^2+1)}{2c^2+b^2+1-2bc}-\left(\frac{2bc+b^2c^2-c^2}{b(c^2+1)} \right)}$$$$\Longrightarrow \frac{\left( \frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} \right) -\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{b(c^2+1)}{2c^2+b^2+1-2bc}-\left(\frac{2bc+b^2c^2-c^2}{b(c^2+1)} \right)}$$$$\Longrightarrow \frac{- \left( \frac{(bc-1)(b^2+2b-c)(b^2-2bc+2c^2+1)+(bc^2+b)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}{(b^3c+b^2-2b^2c^2-3bc+2c^2+1)(b^2-2bc+2c^2+1) } \right) }{\frac{(bc^2+b)^2-(2bc+b^2c^2-c^2)(b^2-2bc+2c^2+1)}{(bc^2+b)(b^2-2bc+2c^2+1)}}$$ $$\Longrightarrow \frac{\frac{-b^5c^2+b^4c^3-2b^4c+3b^3c^2+b^3-b^2c^3-3b^2c+4bc^2+b-2c^3-c}{(b^3c+b^2-2b^2c^2-3bc+2c^2+1)(b^2-2bc+2c^2+1)}}{\frac{-b^4c^2+2b^3c^3-2b^3c-b^2c^4+6b^2c^2+b^2-6bc^3-2bc+2c^4+c^2}{(bc^2+b)(b^2-2bc+2c^2+1)}}$$$$\Longrightarrow \frac{\frac{-b^5c^2+b^4c^3-2b^4c+3b^3c^2+b^3-b^2c^3-3b^2c+4bc^2+b-2c^3-c}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}}{\frac{-b^4c^2+2b^3c^3-2b^3c-b^2c^4+6b^2c^2+b^2-6bc^3-2bc+2c^4+c^2}{bc^2+b}} -----(7)$$\par Now note that, $$\left(-b^5c^2+b^4c^3-2b^4c+3b^3c^2+b^3-b^2c^3-3b^2c+4bc^2+b-2c^3-c \right) \cdot \left(b^3c-2b^2c^2+b^2+3bc^3-2c^4-c^2 \right) $$$$= -b^8c^3+3b^7c^4-3b^7c^2-5b^6c^5+8b^6c^3-b^6c+5b^5c^6-12b^5c^4-2b^5c^2+b^5-2b^4c^7+14b^4c^5+14b^4c^3-2b^4c$$$$-9b^3c^6-24b^3c^4+b^3+2b^2c^7+23b^2c^5+6b^2c^3-b^2c-14bc^6-9bc^4-bc^2+4c^7+4c^5+c^3$$$$=\left( b^4c-b^3c^2+b^3+2b^2c^3+2b^2c-bc^2+b+2c^3+c \right)$$$$ \cdot \left( -b^4c^2+2b^3c^3-2b^3c-b^2c^4+6b^2c^2+b^2-6bc^3-2bc+2c^4+c^2 \right)$$ So using this, ofcourse, $(6)$ and $(7)$ are equal, And Hence, $L$ lies on $KY$ Hence, our Claim is proved!! \par \par \begin{flushleft}

We'll use Cartesian Coordinates to give a proof for this claim! Set $EH_1$ as the $Y- $axis and $DF$ as the $X-$axis Set $D \equiv (c,0)$ and $F \equiv (b,0)$ and WLOG, we can safely assume $E \equiv (0,1)$, because just magnification of the triangle won't change the configuration and won't affect our calculations and Let $b,c \in R$ Since, $H_1$ is the origin, hence, $H_1 \equiv (0,0)$ and really easy to see that the equation of the perpendicular bisector of $DF$ is $x=\frac{b+c}{2}$ $I$ lies on the perpendicular bisector of $DF$: Proof uses the fact that, $BFID$ is cyclic $\Longrightarrow$ $\angle IFD=\angle IDF=\frac{1}{2}\angle ABC$, Using this fact, $I \equiv \left( \frac{b+c}{2} ,y \right) $, we need to calculate the y coordinate of $I$ $\textbf{\underline{Coordinate for I}}$: We know $I\equiv \left( \frac{b+c}{2} , y \right)$, Since $I$ is the circumcenter of $\omega$, therefore, $EI=DI$, hence, by distance formula we get the following: $$EI=DI$$$$\sqrt{\left( \frac{b+c}{2} +0 \right) ^2 +(y-1)^2}=\sqrt{ \left( \frac{b+c}{2} -c \right) ^2 +y^2 }$$And after simplifying this gives $\Longrightarrow$ $y=\frac{bc+1}{2}$, Hence, the coordinate for $I$ is $$I \equiv \left( \frac{b+c}{2} ,\frac{bc+1}{2} \right) $$ We'll use this notation to represent the equation of line throughout the solution: For a line $XY$ with equation $ax+by+c=0$ will be represented as $$XY : ax+by+c=0$$ We'll use the famous formula from coordinate geometry to find out the equation of line when the coordinates for two points on it is given, The formula is as follows: $\textbf{\underline{Formula 1}}$: If $X \equiv (x_1,y_1)$ and $Y \equiv (x_2,y_2)$, then, $$XY: \frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}$$Which can be re-written as: $$XY: \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$$ $\textbf{\underline{Equation for line IE}}:$ We know that the coordinate for $I \equiv \left( \frac{b+c}{2} ,\frac{bc+1}{2} \right)$ and $E \equiv (0,1)$, Hence, $$IE: \frac{y-1}{\frac{bc+1}{2}-1}=\frac{x-0}{\frac{b+c}{2}-0}$$$$IE: y-1=x\left( \frac{bc-1}{b+c} \right)$$ $$IE: y=1+\left( \frac{bc-1}{b+c} \right) x $$ $\textbf{\underline{Equation for line IF}}$: We know that the coordinate for $I \equiv \left(\frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $F \equiv (b,0)$, Hence, $$IF : \frac{y-0}{\frac{bc+1}{2}-0}=\frac{x-b}{\frac{b+c}{2}-b}$$$$IF : y=\left(\frac{x-b}{\frac{c-b}{2}} \right) \left( \frac{bc+1}{2} \right)$$$$IF : y=\left( \frac{bc+1}{c-b} \right) x +\frac{b^2c+b}{b-c}$$ $\textbf{\underline{Equation for line ID}}$: We know that the coordinate for $I \equiv \left( \frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $D \equiv (c,0)$, Hence, $$ID : \frac{y-0}{\frac{bc+1}{2}-0}=\frac{x-c}{\frac{b+c}{2}-c}$$$$ID : y=\left( \frac{x-c}{\frac{b-c}{2}} \right) \left( \frac{bc+1}{2} \right)$$$$ID : y=x \left( \frac{bc+1}{b-c} \right) -\frac{c(bc+1)}{b-c} $$ Note that, $AC \perp EI$, $AB \perp IF$ and $BC \perp ID$ We'll use the famous formula from coordinate geometry to find out the slope of line when the coordinates of two points lying on the line are given, The formula is as follows: $\textbf{\underline{Formula 2}}:$ If $X \equiv (x_1,y_1)$ and $Y \equiv (x_2,y_2)$ and Let $m_{XY}$ represent the slope of the line $XY$, then, $$m_{XY}=\frac{y_1-y_2}{x_1-x_2}$$ $\textbf{\underline{Slope for IF}}$: The coordinate for $I \equiv \left(\frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $F \equiv (b,0)$ $$m_{IF}=\frac{\frac{bc+1}{2}-0}{\frac{b+c}{2}-b}$$$$m_{IF}=\frac{bc+1}{c-b}$$ $\textbf{\underline{Slope for IE}}: $The coordinate for $I \equiv \left( \frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $E \equiv (0,1)$ $$m_{IE}=\frac{\frac{bc+1}{2} -1}{\frac{b+c}{2}-0}$$$$m_{IE}=\frac{bc-1}{b+c}$$ $\textbf{\underline{Slope for ID}}:$ The coordinate for $I \equiv \left( \frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $D \equiv (c,0)$ $$m_{ID}=\frac{\frac{bc+1}{2}-0}{\frac{b+c}{2}-c}$$$$m_{ID}=\frac{bc+1}{b-c}$$ We'll state another famous result from coordinate geometry $\textbf{\underline{Formula 3}}: $If $X_1Y_1$ and $X_2Y_2$ are perpendicular to each other, then, $$m_{X_1Y_1} \cdot m_{X_2Y_2}=-1$$ $\textbf{\underline{Slope for AB}}$: Applying Formula 3 $$m_{AB}=\frac{b-c}{bc+1}$$ $\textbf{\underline{Slope for BC}}:$ Applying Formula 3 $$m_{BC}=\frac{c-b}{bc+1}$$ $ \textbf{\underline{Slope for AC}}$: Applying Formula 3 $$m_{AC}=\frac{-(b+c)}{bc-1}$$ We'll state another well known formula from coordinate geometry to find out the equation of line when the coordinates for a point lying on the line and the slope of line is given $ \textbf{\underline{Formula 4}}$: If $X_1 \equiv (x_1,y_1)$ lies on $\ell$ and $m_{\ell}$ is the slope of $\ell$, then, $$\ell : y-y_1=m_{\ell} (x-x_1)$$ $\textbf{\underline{Equation for line AB}}:$ $F$ lies on $AB$ and $F \equiv (b,0)$ Hence, $$AB: y-0=\left( \frac{b-c}{bc+1} \right) (x-b)$$$$AB: y=\left( \frac{b-c}{bc+1} \right) (x-b)$$ $\textbf{\underline{Equation for line BC}}:$ $D$ lies on $BC$ and $D \equiv (c,0)$ $$BC: y-0=\left( \frac{c-b}{bc+1} \right) (x-c)$$$$BC: y=\left( \frac{c-b}{bc+1} \right) (x-c)$$ $ \textbf{\underline{Equation for line AC}}$: $E$ lies on $AC$ and $E \equiv (0,1)$ $$AC: y-1=\frac{-(b+c)}{bc-1} (x-0)$$$$AC: y=1-x\left( \frac{b+c}{bc-1} \right)$$ $ \textbf{\underline{Coordinate for X}}:$ now recall that $H_1$ is the feet of perpendicular from $E$ to $DF$ and $EH_1 \cap BC=X$ and also, $EH_1$ is the y-axis, and hence, x- coordinate for $X$ is $0$ $\Longrightarrow$ $X \equiv (0,y)$ and since, $X \in BC$, we have, $$BC:y=\left( \frac{c-b}{bc+1} \right) (x-c)$$$$y=-c \left( \frac{c-b}{bc+1} \right)$$$$y=\frac{bc-c^2}{bc+1}$$$$\Longrightarrow X \equiv \left( 0, \frac{bc-c^2}{bc+1} \right)$$ $\textbf{\underline{Equation for line DE}}:$ Recall, $D \equiv (c,0)$ and $E \equiv (0,1)$, Hence, using Formula 1, $$DE:\frac{y-0}{1-0}=\frac{x-c}{0-c}$$$$DE: y=1-\frac{x}{c}$$ $ \textbf{\underline{Equation for line EF}}:$ Recall, $E \equiv (0,1)$ and $F \equiv (b,0)$, Hence, using Formula 1, $$EF: \frac{y-0}{1-0}=\frac{x-b}{0-b}$$$$EF: y=1-\frac{x}{b}$$ $ \textbf{\underline{Slope for DE}}$: $D \equiv (c,0)$ and $E \equiv (0,1)$, $$m_{DE}=\frac{0-1}{c-0}$$$$m_{DE}=\frac{-1}{c}$$ $\textbf{\underline{Equation for line $FH_2$}}$: $F \equiv (b,0)$ Notice, $FH_2 \perp ED$, hence, $$m_{FH_2} \cdot m_{DE}=-1$$$$m_{FH_2}=c$$And since, $F$ lies on $FH_2$, therefore, $$FH_2 : y-0=c(x-b)$$$$FH_2 : y=c(x-b)$$ $ \textbf{\underline{Coordinate for Y}}$: Recall $FH_2 \cap BC=Y$, we have, $FH_2 : y=c(x-b)$ and $BC: y=\left( \frac{c-b}{bc+1} \right) (x-c)$, Hence, solving equations: should yield coordinates for $Y$ $$FH_2 : y=c(x-b) -----(1)$$$$BC: y=\left( \frac{c-b}{bc+1} \right) (x-c) ------(2)$$ Equating $(1)$ and $(2)$, we have, $$c(x-b)=\frac{(c-b)(x-c)}{bc+1}$$$$\frac{1}{bc+1}=\frac{c(x-b)}{(c-b)(x-c)}=\frac{cx-bc}{cx+bc-bx-c^2}$$$$cx+bc-bx-c^2=(cx-bc)(bc+1)$$$$cx+bc-bx-c^2=bc^2x+cx-b^2c^2-bc$$$$2bc-c^2-bx=bc^2x-b^2c^2$$$$2bc-c^2+b^2c^2=bc^2x+bx$$$$x=\frac{2bc-c^2+b^2c^2}{b(c^2+1)}$$ and, $$y=c(x-b)$$$$y=c \left( \frac{2bc-c^2+b^2c^2}{b(c^2+1)} -b \right)$$$$y=c \left( \frac{2bc-c^2+b^2c^2-b^2c^2-b^2}{bc^2+b} \right)$$$$y=\frac{-c(b-c)^2}{bc^2+b}$$ Hence, $$Y \equiv \left( \frac{2bc+b^2c^2-c^2}{b(c^2+1)} , \frac{-c(b-c)^2}{b(c^2+1)} \right)$$ Let $D^*$ be a point diametrically opposite to $D$ on $\omega$, or inshort, Let $D^*$ be the $D-$antipode in $\omega$, hence, $\angle D^*FD=\angle EH_1D=90^{\circ}$ $\Longrightarrow$ $FD^*||EH_1$ $$\Longrightarrow D^* \equiv (b,y)$$and since, $D^* \in ID$ and $ID:y=x \left( \frac{bc+1}{b-c} \right) -\frac{c(bc+1)}{b-c} $ $$y=x \left( \frac{bc+1}{b-c} \right) -\frac{c(bc+1)}{b-c} $$$$y=(x-c) \left( \frac{bc+1}{b-c} \right) $$$$y=(b-c) \left( \frac{bc+1}{b-c} \right)$$$$y=bc+1$$Hence, $$D^* \equiv (b,bc+1)$$ $\textbf{\underline{Coordinate for A}}$:Recall, $AB \cap AC=A$ $$AB: y=\left( \frac{b-c}{bc+1} \right) (x-b)$$$$AC:y=1-x\left( \frac{b+c}{bc-1} \right)$$ $$\left( \frac{b-c}{bc+1} \right) (x-b)=1-x\left( \frac{b+c}{bc-1} \right)$$$$\frac{bc-1}{bc+1}=\frac{(bc-1)-x(b+c)}{(b-c)(x-b)}$$$$(b-c)(x-b)(bc-1)=(bc+1)(bc-1)-x(b+c)(bc+1)$$$$(b^2c-bc^2-b+c)(x-b)=b^2c^2-1-(b^2c+b+bc^2+c)x$$$$b^2c^2-1-(b^2c+b+bc^2+c)x=x(b^2c-bc^2-b+c) -b(b^2c-bc^2-b+c)$$$$b^2c^2-1+b(b^2c-bc^2-b+c)=x(b^2c-bc^2-b+c+b^2c+b+c+bc^2)$$$$b^2c^2-1+b^3c-b^2c^2-b^2+bc=x(2b^2c+2c)$$$$b^3c-b^2+bc-1=x(2b^2c+2c)$$$$(bc-1)(b^2+1)=x(2c)(b^2+1)$$$$x=\frac{bc-1}{2c}$$And, $$y=1-x\left( \frac{b+c}{bc-1} \right)$$$$y=1-\frac{b+c}{2c}$$$$y=\frac{c-b}{2c}$$Hence, $$A \equiv \left( \frac{bc-1}{2c} , \frac{c-b}{2c} \right) $$ $\textbf{\underline{Equation for line AD}}:$ Recall, $D \equiv (c,0)$ $$AD: \frac{y-0}{\frac{c-b}{2c}-0}=\frac{x-c}{\frac{bc-1}{2c}-c}$$$$AD: y=\frac{x-c}{\frac{bc-1-2c^2}{2c}} \left(\frac{c-b}{2c} \right)$$$$AD: y =\frac{(x-c)(c-b)}{bc-1-2c^2}$$ Recall $AZ \cap EF=K$, but since, $Z \in AD$, hence, $AD \cap EF=K$ $\textbf{\underline{Coordinate for K}}$: Recall, $AD \cap EF=K$, We have, $$AD: y =\frac{(x-c)(c-b)}{bc-1-2c^2}$$$$EF: y=1-\frac{x}{b}$$ Hence, $$1-\frac{x}{b}=\frac{(x-c)(c-b)}{bc-1-2c^2}$$ $$1-\frac{x}{b}=\frac{x(c-b)}{bc-1-2c^2} +\frac{c(b-c)}{bc-1-2c^2}$$$$1-\frac{(bc-c^2)}{bc-1-12c^2}=\frac{x}{b}+\frac{x(c-b)}{bc-1-2c^2}$$$$\frac{bc-1-2c^2-bc+c^2}{bc-1-2c^2}=\frac{x(bc-1-2c^2+bc-b^2)}{b(bc-1-2c^2)}$$$$\frac{-(c^2+1)}{bc-1-2c^2}=\frac{x(2bc-1-2c^2-b^2)}{b(bc-1-2c^2}$$$$x=\frac{-b(c^2+1)}{2bc-1-2c^2-b^2}$$$$x=\frac{b(c^2+1)}{2c^2+b^2+1-2bc}$$And, $$y=1-\frac{x}{b}$$$$y=1-\frac{b(c^2+1)}{b(2c^2+b^2+1-2bc)}$$$$y=\frac{2c^2+b^2+1-2bc-c^2-1}{2c^2+b^2+1-2bc}$$$$y=\frac{(b-c)^2}{2c^2+b^2+1-2bc}$$Hence, $$K \equiv \left( \frac{b(c^2+1)}{2c^2+b^2+1-2bc} , \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)$$ $\textbf{\underline{Coodinate for Z}}$: Note, that, since, $DD^*$ is the diameter of $\omega$, hence, $ZD^* \perp ZD$ or $ZD^* \perp AD$, therefore, $$m_{ZD^*} \cdot m_{AD}=-1$$Now we have eariler found out that, $$AD: y=\frac{(x-c)(c-b)}{bc-1-2c^2}$$$$AD:y=\frac{x(c-b)}{bc-1-2c^2} -\frac{c(b-c)}{bc-1-2c^2}$$We know that equation of a line can be represented in the form of $y=mx+b$, where $m$ is the slope of the line and $b$ is the $y-$intercept, hence using this we get, $$m_{AD}=\frac{(c-b)}{bc-1-2c^2}$$ Hence, $$ m_{ZD^*}=\frac{-1}{m_{AD}}=\frac{-1}{\frac{c-b}{bc-1-2c^2}}$$ $$m_{ZD^*}=\frac{2c^2+1-bc}{c-b}$$Let $Z \equiv (x',y')$ and recall that $D^* \equiv (b,bc+1)$, hence ,we can find equation of $ZD^*$, $$ZD^*: \frac{y-y'}{bc+1-y'}=\frac{x-x'}{b-x'}$$$$ZD^*:y=y'+\left( \frac{x-x'}{b-x'} \right) (bc+1-y')$$Using the earier we stated that, any line can be written in the form of $y=mx+b$, so using this, $$m_{ZD^*}=\frac{bc+1-y'}{b-x'}$$but, we eariler found out that, $$m_{ZD^*}=\frac{2c^2+1-bc}{c-b}$$$$\Longrightarrow \frac{2c^2+1-bc}{c-b}=\frac{bc+1-y'}{b-x'} ----(3)$$And, Since, $Z \in AD$, $$AD: y=\frac{(x-c)(b-c)}{bc-1-2c^2}$$$$y'=\frac{(x-c)(b-c)}{bc-1-2c^2}----(4)$$Solving $(3)$ and $(4)$ should yield the coordinates for $Z$, $$(3) \longrightarrow$$$$\frac{2c^2+1-bc}{c-b}=\frac{bc+1-y'}{b-x'}$$$$(2c^2+1-bc)(b-x')=(bc+1)(c-b)-y'(c-b)$$$$y'=\frac{(2c^2+1-bc)(b-x')+(bc+1)(b-c)}{b-c} -----(5)$$ Therefore, dividing $(4)$ and $(5)$, we have, $$\frac{y'}{y'}=\frac{\frac{(x'-c)(c-b)}{bc-1-2c^2}}{\frac{(2c^2+1-bc)(b-x')+(bc+1)(b-c)}{b-c}}$$$$\frac{bc-1-2c^2}{b-c}=\frac{(x'-c)(c-b)}{(2c^2+1-bc)(b-x')+(bc+1)(b-c)}$$$$\frac{bc-1-2c^2}{b-c}=\frac{x'(c-b) +bc-c^2}{2bc^2+b-b^2c-x'(2c^2+1-bc)+b^2c+b-bc^2-c}$$$$\frac{bc-1-2c^2}{b-c}=\frac{x'(c-b) +bc-c^2}{bc^2+2b-c-x'(2c^2+1-bc)}$$$$(bc-1-2c^2)(bc^2+2b-c-x'(2c^2+1-bc))=x'(c-b)(b-c)+(bc-c^2)(b-c)$$$$(bc-1-2c^2)(bc^2+2b-c)+x'(bc-1-2c^2)^2=c(b-c)^2-x'(b-c)^2$$$$b^2c^3+2b^2c-bc^2-bc^2-2b+c-2bc^4-4bc^2+2c^3+x'(bc-1-2c^2)^2=b^2c-2bc^2+c^3-x'(b-c)^2$$$$b^2c^3+2b^2c-bc^2-bc^2-2b+c-2bc^4-4bc^2+2c^3-b^2c+2bc^2-c^3=-x'((b-c)^2+(bc-1-2c^2))$$$$(b^2c-2bc^2-2b+c)(c^2+1)=-x'(b^2-2bc+c^2+b^2c^2+1+4c^4-2bc+4c^2-4bc^3)$$$$(b^2c-2bc^2-2b+c)(c^2+1)=-x'((c^2+1)(b^2-4bc+4c^2+1))$$$$x' \equiv - \left( \frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} \right)$$ And, Since, as above stated, $$y'=\frac{(x'-c)(c-b)}{bc-1-2c^2}$$$$y'=\frac{(-(\frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1})-c)(c-b)}{bc-1-2c^2}$$$$y'=\frac{-(\frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} +c)(c-b)}{bc-1-2c^2}$$$$y'=\frac{(\frac{b^2c-2bc^2-2b+c+b^2c-4bc^2+4c^3+c}{b^2-4bc+4c^2+1})(b-c)}{bc-1-2c^2}$$$$y'=\frac{2(b-c)(bc-1-2c^2)(b-c)}{(bc-1-2c^2)(b^2-4bc+4c^2+1)}$$$$y' \equiv \frac{2(b-c)^2}{b^2-4bc+4c^2+1}$$Hence, Combining we have the coordinates for $Z$ $$Z \equiv \left( - \left( \frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} \right) , \left( \frac{2(b-c)^2}{b^2-4bc+4c^2+1} \right) \right)$$ $\textbf{\underline{Equation for line $ZD^*$ }}$: Recall, $D^* \equiv (b,bc+1)$ and we have found out the coordinate for $Z$ above, Hence, $$ZD^* : \frac{y-(bc+1)}{\frac{2(b-c)^2}{b^2-4bc+4c^2+1} -(bc+1)}=\frac{x-b}{- \left( \frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} \right) -b}$$$$ZD^*: \frac{y-(bc+1)}{2(b-c)^2-(bc+1)(b^2-4bc+4c^2+1)}=\frac{x-b}{-(b^2c-2bc^2-2b+c+b^3-4b^2c+4bc^2+b)}$$$$ZD^*: \frac{y-(bc+1)}{-(b^2-2bc-1)(bc-2c^2-1)}=\frac{x-b}{-(b-c)(b^2-2bc-1)}$$ $$ZD^*: y=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$ $\textbf{\underline{Equation for line AX}}:$ Recall, $A \equiv \left( \frac{bc-1}{2c} , \frac{c-b}{2c} \right)$ and $X \equiv \left( 0, \frac{bc-c^2}{bc+1} \right) $, Hence, $$AX: \frac{y- \frac{c-b}{2c}}{\frac{bc-c^2}{bc+1}-(\frac{c-b}{2c})}=\frac{x-\frac{bc-1}{2c}}{0-(\frac{bc-1}{2c})}$$$$AX: \frac{y-\frac{c-b}{2c}}{\frac{(bc-c^2)(2c)-(c-b)(bc+1)}{(bc+1)(2c)}} =\frac{x-(\frac{bc-1}{2c})}{-(\frac{bc-1}{2c})}$$$$AX: \frac{y-(\frac{c-b}{2c})}{\frac{b^2c+bc^2+b-2c^3-c}{bc+1}}=\frac{x-(\frac{bc-1}{2c})}{1-bc}$$$$AX: y-\left(\frac{c-b}{2c} \right)=\left( \frac{2cx-(bc-1)}{2c(1-bc)} \right) \left( \frac{b^2c+bc^2+b-2c^3-c)}{bc+1} \right)$$ $$AX: y=\frac{c-b}{2c}- \left( \frac{(2cx-(bc-1))(b^2c+bc^2+b-2c^3-c)}{2c(b^2c^2-1)} \right)$$ $\textbf{\underline{Coordinate for L}}$: Recall, that, $AX \cap ZD^*=L$ We have already found the equation for line $AX$ and line $ZD^*$, so now, we can find out the coordinates for $L$, $$AX: y=\frac{c-b}{2c}- \left( \frac{(2cx-(bc-1))(b^2c+bc^2+b-2c^3-c)}{2c(b^2c^2-1)} \right)$$$$ZD^*: y=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$Equating $y$, we get, $$\Longrightarrow \frac{c-b}{2c}- \left( \frac{(2cx-(bc-1))(b^2c+bc^2+b-2c^3-c)}{2c(b^2c^2-1)} \right)=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$$$\Longrightarrow -(b-c)^2(b^2c^2-1)-(2cx+1-bc)(b^2c+bc^2+b-2c^3-c)(b-c)$$$$=2c(bc+1)(b-c)(b^2c^2-1)+(2c)(bc-1-2c^2)(x-b)(b^2c^2-1)$$ $$\Longrightarrow -2b^3c^4x+4b^2c^5x-6c^3x-2cx-2b^3c^2x-2b^2cx+6bc^4x+6bc^2x-4c^5x+2b^2c^3x$$$$=2b^3c^3+2c^2+2b^3c^5-4bc+2b^2c^4+2b^2c^2-2bc^5-6bc^3+2c^4$$ $$\Longrightarrow -2c(c^2+1)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)x=2c(c^2+1)(bc-1)(b^2c+2b-c)$$$$\Longrightarrow x=\frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}$$ Now, we can use the value of $x$ to find out the value of $y$, $$y=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$$$y=\frac{(bc+1)(b-c) + (bc-2c^2-1)(x-b)}{b-c}$$$$y=\frac{(b^2-bc^2+b-c)+(bc-2c^2-1) \left(\frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}-b \right)}{b-c}$$$$y=\frac{(bc+1)(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)-(bc-2c^2-1)(b^4c-b^3c^2+b^3-2b^2c+bc^2-b+c)}{(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}$$$$y=\frac{2b^4c-4b^3c^2+2b^3+2b^2c^3-4b^2c+2bc^2}{(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}$$$$y=\frac{2b(b-c)^2(bc+1)}{(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}$$$$y=\frac{2b(b-c)(bc+1)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}$$ $$L \equiv \left( \frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} , \frac{2b(b-c)(bc+1)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} \right)$$ Now we'll calculate the equation for $YK$ $\textbf{\underline{Equation for line YK}}$: Recall, $Y \equiv \left( \frac{2bc+b^2c^2-c^2}{b(c^2+1)} , \frac{-c(b-c)^2}{b(c^2+1)} \right)$ and \par $K \equiv \left( \frac{b(c^2+1)}{2c^2+b^2+1-2bc} , \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)$ Hence, we have, $$YK : \frac{y-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{-c(b-c)^2}{b(c^2+1)}-\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}=\frac{x-\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{2bc+b^2c^2-c^2}{b(c^2+1)}-\left(\frac{b(c^2+1)}{2c^2+b^2+1-2bc}\right)} $$$$YK : \frac{y-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{c(b-c)^2}{b(c^2+1)}+\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}=\frac{x-\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{b(c^2+1)}{2c^2+b^2+1-2bc}-\left(\frac{2bc+b^2c^2-c^2}{b(c^2+1)} \right)} $$ Now if the coordinates of $L$ satisfy equation of $YK$, then we have indeed proved that $L$ lies on $YK$, so we'll do it by first computing the RHS and then LHS and then if RHS equals LHS, we're done with our claim, So here we go!, $\textbf{\underline{LHS}}:$ $$\Longrightarrow \frac{y-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{c(b-c)^2}{b(c^2+1)}+\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}$$$$\Longrightarrow \frac{\left( \frac{2b(b-c)(bc+1)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} \right)-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{c(b-c)^2}{b(c^2+1)}+\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}$$$$\Longrightarrow \frac{\frac{(b^2-2bc+2c^2+1)(2b^2-2bc)(bc+1)-(b-c)^2(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}{(b^3c+b^2-2b^2c^2-3bc+2c^2+1)(b^2-2bc+2c^2+1)}}{\frac{c(b-c)^2(b^2-2bc+2c^2+1)+(b-c)^2(bc^2+b)}{(bc^2+b)(b^2-2bc+2c^2+1)}}$$$$\Longrightarrow \frac{\frac{b^4c-b^3c^2+b^3+2b^2c^3+2b^2c-bc^2+b+2c^3+c}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}}{\frac{(b-c)(b^2c-bc^2+b+2c^3+c)}{bc^2+b}}$$$$\Longrightarrow \frac{\frac{b^4c-b^3c^2+b^3+2b^2c^3+2b^2c-bc^2+b+2c^3+c}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}}{\frac{b^3c-2b^2c^2+b^2+3bc^3-2c^4-c^2}{bc^2+b}} ------(6)$$ $\textbf{\underline{RHS}}: $ $$\Longrightarrow \frac{x-\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{b(c^2+1)}{2c^2+b^2+1-2bc}-\left(\frac{2bc+b^2c^2-c^2}{b(c^2+1)} \right)}$$$$\Longrightarrow \frac{\left( \frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} \right) -\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{b(c^2+1)}{2c^2+b^2+1-2bc}-\left(\frac{2bc+b^2c^2-c^2}{b(c^2+1)} \right)}$$$$\Longrightarrow \frac{- \left( \frac{(bc-1)(b^2+2b-c)(b^2-2bc+2c^2+1)+(bc^2+b)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}{(b^3c+b^2-2b^2c^2-3bc+2c^2+1)(b^2-2bc+2c^2+1) } \right) }{\frac{(bc^2+b)^2-(2bc+b^2c^2-c^2)(b^2-2bc+2c^2+1)}{(bc^2+b)(b^2-2bc+2c^2+1)}}$$ $$\Longrightarrow \frac{\frac{-b^5c^2+b^4c^3-2b^4c+3b^3c^2+b^3-b^2c^3-3b^2c+4bc^2+b-2c^3-c}{(b^3c+b^2-2b^2c^2-3bc+2c^2+1)(b^2-2bc+2c^2+1)}}{\frac{-b^4c^2+2b^3c^3-2b^3c-b^2c^4+6b^2c^2+b^2-6bc^3-2bc+2c^4+c^2}{(bc^2+b)(b^2-2bc+2c^2+1)}}$$$$\Longrightarrow \frac{\frac{-b^5c^2+b^4c^3-2b^4c+3b^3c^2+b^3-b^2c^3-3b^2c+4bc^2+b-2c^3-c}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}}{\frac{-b^4c^2+2b^3c^3-2b^3c-b^2c^4+6b^2c^2+b^2-6bc^3-2bc+2c^4+c^2}{bc^2+b}} -----(7)$$ Now note that, $$\left(-b^5c^2+b^4c^3-2b^4c+3b^3c^2+b^3-b^2c^3-3b^2c+4bc^2+b-2c^3-c \right) \cdot \left(b^3c-2b^2c^2+b^2+3bc^3-2c^4-c^2 \right) $$$$= -b^8c^3+3b^7c^4-3b^7c^2-5b^6c^5+8b^6c^3-b^6c+5b^5c^6-12b^5c^4-2b^5c^2+b^5-2b^4c^7+14b^4c^5+14b^4c^3-2b^4c$$$$-9b^3c^6-24b^3c^4+b^3+2b^2c^7+23b^2c^5+6b^2c^3-b^2c-14bc^6-9bc^4-bc^2+4c^7+4c^5+c^3$$$$=\left( b^4c-b^3c^2+b^3+2b^2c^3+2b^2c-bc^2+b+2c^3+c \right)$$$$ \cdot \left( -b^4c^2+2b^3c^3-2b^3c-b^2c^4+6b^2c^2+b^2-6bc^3-2bc+2c^4+c^2 \right)$$ So using this, ofcourse, $(6)$ and $(7)$ are equal, And Hence, $L$ lies on $KY$ Hence, our Claim is proved!! \begin{flushleft}

Draw $\ell$ parallel to $XC$ through $Y$, then $CY \perp \ell$ and let $\ell \cap AC=P$ and $\ell \cap BC=Q$, then, $\Delta XAC \sim \Delta YAP$ and $\Delta XBC \sim \Delta YBQ$ gives us, $$\frac{AX}{XC}=\frac{AY}{PY} -----(i)$$And, $$\frac{BX}{XC}=\frac{BY}{YQ} -----(ii)$$and, $$-1=(X,Y;A,B) \Longrightarrow \frac{AX}{AY}=\frac{BX}{BY} ------(iii)$$Hence, using $(i), (ii), (iii)$, it's easy to get, $$YP=YQ \Longrightarrow \angle ACY=\angle BCA$$Hence, $CY$ is the angle bisector of $\angle ACB$ Hence we're completed with the proof of our Lemma!!

We'll use Cartesian Coordinates to give a proof for this claim! Set $EH_1$ as the $Y- $axis and $DF$ as the $X-$axis Set $D \equiv (c,0)$ and $F \equiv (b,0)$ and WLOG, we can safely assume $E \equiv (0,1)$, because just magnification of the triangle won't change the configuration and won't affect our calculations and Let $b,c \in R$ Since, $H_1$ is the origin, hence, $H_1 \equiv (0,0)$ and really easy to see that the equation of the perpendicular bisector of $DF$ is $x=\frac{b+c}{2}$ $I$ lies on the perpendicular bisector of $DF$: Proof uses the fact that, $BFID$ is cyclic $\Longrightarrow$ $\angle IFD=\angle IDF=\frac{1}{2}\angle ABC$, Using this fact, $I \equiv \left( \frac{b+c}{2} ,y \right) $, we need to calculate the y coordinate of $I$ $\textbf{\underline{Coordinate for I}}$: We know $I\equiv \left( \frac{b+c}{2} , y \right)$, Since $I$ is the circumcenter of $\omega$, therefore, $EI=DI$, hence, by distance formula we get the following: $$EI=DI$$$$\sqrt{\left( \frac{b+c}{2} +0 \right) ^2 +(y-1)^2}=\sqrt{ \left( \frac{b+c}{2} -c \right) ^2 +y^2 }$$And after simplifying this gives $\Longrightarrow$ $y=\frac{bc+1}{2}$, Hence, the coordinate for $I$ is $$I \equiv \left( \frac{b+c}{2} ,\frac{bc+1}{2} \right) $$ We'll use this notation to represent the equation of line throughout the solution: For a line $XY$ with equation $ax+by+c=0$ will be represented as $$XY : ax+by+c=0$$ \end{flushleft} \begin{flushleft} We'll use the famous formula from coordinate geometry to find out the equation of line when the coordinates for two points on it is given, The formula is as follows: \par \end{flushleft} \par \begin{flushleft} \textbf{\underline{Formula 1}}: If $X \equiv (x_1,y_1)$ and $Y \equiv (x_2,y_2)$, then, $$XY: \frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}$$Which can be re-written as: $$XY: \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$$ \end{flushleft} \par \textbf{\underline{Equation for line IE}}: We know that the coordinate for $I \equiv \left( \frac{b+c}{2} ,\frac{bc+1}{2} \right)$ and $E \equiv (0,1)$, Hence, $$IE: \frac{y-1}{\frac{bc+1}{2}-1}=\frac{x-0}{\frac{b+c}{2}-0}$$$$IE: y-1=x\left( \frac{bc-1}{b+c} \right)$$ $$IE: y=1+\left( \frac{bc-1}{b+c} \right) x $$ \par \textbf{\underline{Equation for line IF}}: We know that the coordinate for $I \equiv \left(\frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $F \equiv (b,0)$, Hence, $$IF : \frac{y-0}{\frac{bc+1}{2}-0}=\frac{x-b}{\frac{b+c}{2}-b}$$$$IF : y=\left(\frac{x-b}{\frac{c-b}{2}} \right) \left( \frac{bc+1}{2} \right)$$$$IF : y=\left( \frac{bc+1}{c-b} \right) x +\frac{b^2c+b}{b-c}$$ \textbf{\underline{Equation for line ID}}: We know that the coordinate for $I \equiv \left( \frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $D \equiv (c,0)$, Hence, $$ID : \frac{y-0}{\frac{bc+1}{2}-0}=\frac{x-c}{\frac{b+c}{2}-c}$$$$ID : y=\left( \frac{x-c}{\frac{b-c}{2}} \right) \left( \frac{bc+1}{2} \right)$$$$ID : y=x \left( \frac{bc+1}{b-c} \right) -\frac{c(bc+1)}{b-c} $$ \par \begin{flushleft} Note that, $AC \perp EI$, $AB \perp IF$ and $BC \perp ID$ \end{flushleft} \par \begin{flushleft} We'll use the famous formula from coordinate geometry to find out the slope of line when the coordinates of two points lying on the line are given, The formula is as follows: \par \end{flushleft} \par \begin{flushleft} \textbf{\underline{Formula 2}}: If $X \equiv (x_1,y_1)$ and $Y \equiv (x_2,y_2)$ and Let $m_{XY}$ represent the slope of the line $XY$, then, $$m_{XY}=\frac{y_1-y_2}{x_1-x_2}$$\end{flushleft} \textbf{\underline{Slope for IF}}: The coordinate for $I \equiv \left(\frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $F \equiv (b,0)$ $$m_{IF}=\frac{\frac{bc+1}{2}-0}{\frac{b+c}{2}-b}$$$$m_{IF}=\frac{bc+1}{c-b}$$ \textbf{\underline{Slope for IE}}: The coordinate for $I \equiv \left( \frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $E \equiv (0,1)$ $$m_{IE}=\frac{\frac{bc+1}{2} -1}{\frac{b+c}{2}-0}$$$$m_{IE}=\frac{bc-1}{b+c}$$ \textbf{\underline{Slope for ID}}: The coordinate for $I \equiv \left( \frac{b+c}{2} , \frac{bc+1}{2} \right)$ and $D \equiv (c,0)$ $$m_{ID}=\frac{\frac{bc+1}{2}-0}{\frac{b+c}{2}-c}$$$$m_{ID}=\frac{bc+1}{b-c}$$ \begin{flushleft} We'll state another famous result from coordinate geometry \par \end{flushleft} \par \begin{flushleft} \textbf{\underline{Formula 3}}: If $X_1Y_1$ and $X_2Y_2$ are perpendicular to each other, then, $$m_{X_1Y_1} \cdot m_{X_2Y_2}=-1$$\par \end{flushleft} \par \textbf{\underline{Slope for AB}}: Applying Formula 3 $$m_{AB}=\frac{b-c}{bc+1}$$ \textbf{\underline{Slope for BC}}: Applying Formula 3 $$m_{BC}=\frac{c-b}{bc+1}$$ \textbf{\underline{Slope for AC}}: Applying Formula 3 $$m_{AC}=\frac{-(b+c)}{bc-1}$$ \begin{flushleft} We'll state another well known formula from coordinate geometry to find out the equation of line when the coordinates for a point lying on the line and the slope of line is given \par \end{flushleft} \par \begin{flushleft} \textbf{\underline{Formula 4}}: If $X_1 \equiv (x_1,y_1)$ lies on $\ell$ and $m_{\ell}$ is the slope of $\ell$, then, $$\ell : y-y_1=m_{\ell} (x-x_1)$$\end{flushleft} \textbf{\underline{Equation for line AB}}: $F$ lies on $AB$ and $F \equiv (b,0)$ Hence, $$AB: y-0=\left( \frac{b-c}{bc+1} \right) (x-b)$$$$AB: y=\left( \frac{b-c}{bc+1} \right) (x-b)$$ \textbf{\underline{Equation for line BC}}: $D$ lies on $BC$ and $D \equiv (c,0)$ $$BC: y-0=\left( \frac{c-b}{bc+1} \right) (x-c)$$$$BC: y=\left( \frac{c-b}{bc+1} \right) (x-c)$$ \textbf{\underline{Equation for line AC}}: $E$ lies on $AC$ and $E \equiv (0,1)$ $$AC: y-1=\frac{-(b+c)}{bc-1} (x-0)$$$$AC: y=1-x\left( \frac{b+c}{bc-1} \right)$$ \textbf{\underline{Coordinate for X}}: now recall that $H_1$ is the feet of perpendicular from $E$ to $DF$ and $EH_1 \cap BC=X$ and also, $EH_1$ is the y-axis, and hence, x- coordinate for $X$ is $0$ $\Longrightarrow$ $X \equiv (0,y)$ and since, $X \in BC$, we have, $$BC:y=\left( \frac{c-b}{bc+1} \right) (x-c)$$$$y=-c \left( \frac{c-b}{bc+1} \right)$$$$y=\frac{bc-c^2}{bc+1}$$$$\Longrightarrow X \equiv \left( 0, \frac{bc-c^2}{bc+1} \right)$$ \textbf{\underline{Equation for line DE}}: Recall, $D \equiv (c,0)$ and $E \equiv (0,1)$, Hence, using Formula 1, $$DE:\frac{y-0}{1-0}=\frac{x-c}{0-c}$$$$DE: y=1-\frac{x}{c}$$ \textbf{\underline{Equation for line EF}}: Recall, $E \equiv (0,1)$ and $F \equiv (b,0)$, Hence, using Formula 1, $$EF: \frac{y-0}{1-0}=\frac{x-b}{0-b}$$$$EF: y=1-\frac{x}{b}$$ \textbf{\underline{Slope for DE}}: $D \equiv (c,0)$ and $E \equiv (0,1)$, $$m_{DE}=\frac{0-1}{c-0}$$$$m_{DE}=\frac{-1}{c}$$ \textbf{\underline{Equation for line $FH_2$}}: $F \equiv (b,0)$ Notice, $FH_2 \perp ED$, hence, $$m_{FH_2} \cdot m_{DE}=-1$$$$m_{FH_2}=c$$And since, $F$ lies on $FH_2$, therefore, $$FH_2 : y-0=c(x-b)$$$$FH_2 : y=c(x-b)$$ \textbf{\underline{Coordinate for Y}}: Recall $FH_2 \cap BC=Y$, we have, $FH_2 : y=c(x-b)$ and $BC: y=\left( \frac{c-b}{bc+1} \right) (x-c)$, Hence, solving equations: should yield coordinates for $Y$ $$FH_2 : y=c(x-b) -----(1)$$$$BC: y=\left( \frac{c-b}{bc+1} \right) (x-c) ------(2)$$ Equating $(1)$ and $(2)$, we have, $$c(x-b)=\frac{(c-b)(x-c)}{bc+1}$$$$\frac{1}{bc+1}=\frac{c(x-b)}{(c-b)(x-c)}=\frac{cx-bc}{cx+bc-bx-c^2}$$$$cx+bc-bx-c^2=(cx-bc)(bc+1)$$$$cx+bc-bx-c^2=bc^2x+cx-b^2c^2-bc$$$$2bc-c^2-bx=bc^2x-b^2c^2$$$$2bc-c^2+b^2c^2=bc^2x+bx$$$$x=\frac{2bc-c^2+b^2c^2}{b(c^2+1)}$$ and, $$y=c(x-b)$$$$y=c \left( \frac{2bc-c^2+b^2c^2}{b(c^2+1)} -b \right)$$$$y=c \left( \frac{2bc-c^2+b^2c^2-b^2c^2-b^2}{bc^2+b} \right)$$$$y=\frac{-c(b-c)^2}{bc^2+b}$$ Hence, $$Y \equiv \left( \frac{2bc+b^2c^2-c^2}{b(c^2+1)} , \frac{-c(b-c)^2}{b(c^2+1)} \right)$$ \par \begin{flushleft} Let $D^*$ be a point diametrically opposite to $D$ on $\omega$, or inshort, Let $D^*$ be the $D-$antipode in $\omega$, hence, $\angle D^*FD=\angle EH_1D=90^{\circ}$ $\Longrightarrow$ $FD^*||EH_1$ $$\Longrightarrow D^* \equiv (b,y)$$and since, $D^* \in ID$ and $ID:y=x \left( \frac{bc+1}{b-c} \right) -\frac{c(bc+1)}{b-c} $ $$y=x \left( \frac{bc+1}{b-c} \right) -\frac{c(bc+1)}{b-c} $$$$y=(x-c) \left( \frac{bc+1}{b-c} \right) $$$$y=(b-c) \left( \frac{bc+1}{b-c} \right)$$$$y=bc+1$$Hence, $$D^* \equiv (b,bc+1)$$\par \end{flushleft} \par \textbf{\underline{Coordinate for A}}:Recall, $AB \cap AC=A$ $$AB: y=\left( \frac{b-c}{bc+1} \right) (x-b)$$$$AC:y=1-x\left( \frac{b+c}{bc-1} \right)$$ $$\left( \frac{b-c}{bc+1} \right) (x-b)=1-x\left( \frac{b+c}{bc-1} \right)$$$$\frac{bc-1}{bc+1}=\frac{(bc-1)-x(b+c)}{(b-c)(x-b)}$$$$(b-c)(x-b)(bc-1)=(bc+1)(bc-1)-x(b+c)(bc+1)$$$$(b^2c-bc^2-b+c)(x-b)=b^2c^2-1-(b^2c+b+bc^2+c)x$$$$b^2c^2-1-(b^2c+b+bc^2+c)x=x(b^2c-bc^2-b+c) -b(b^2c-bc^2-b+c)$$$$b^2c^2-1+b(b^2c-bc^2-b+c)=x(b^2c-bc^2-b+c+b^2c+b+c+bc^2)$$$$b^2c^2-1+b^3c-b^2c^2-b^2+bc=x(2b^2c+2c)$$$$b^3c-b^2+bc-1=x(2b^2c+2c)$$$$(bc-1)(b^2+1)=x(2c)(b^2+1)$$$$x=\frac{bc-1}{2c}$$And, $$y=1-x\left( \frac{b+c}{bc-1} \right)$$$$y=1-\frac{b+c}{2c}$$$$y=\frac{c-b}{2c}$$Hence, $$A \equiv \left( \frac{bc-1}{2c} , \frac{c-b}{2c} \right) $$ \textbf{\underline{Equation for line AD}}: Recall, $D \equiv (c,0)$ $$AD: \frac{y-0}{\frac{c-b}{2c}-0}=\frac{x-c}{\frac{bc-1}{2c}-c}$$$$AD: y=\frac{x-c}{\frac{bc-1-2c^2}{2c}} \left(\frac{c-b}{2c} \right)$$$$AD: y =\frac{(x-c)(c-b)}{bc-1-2c^2}$$\par \begin{flushleft} Recall $AZ \cap EF=K$, but since, $Z \in AD$, hence, $AD \cap EF=K$ \par \end{flushleft} \par \textbf{\underline{Coordinate for K}}: Recall, $AD \cap EF=K$, We have, $$AD: y =\frac{(x-c)(c-b)}{bc-1-2c^2}$$$$EF: y=1-\frac{x}{b}$$ Hence, $$1-\frac{x}{b}=\frac{(x-c)(c-b)}{bc-1-2c^2}$$ $$1-\frac{x}{b}=\frac{x(c-b)}{bc-1-2c^2} +\frac{c(b-c)}{bc-1-2c^2}$$$$1-\frac{(bc-c^2)}{bc-1-12c^2}=\frac{x}{b}+\frac{x(c-b)}{bc-1-2c^2}$$$$\frac{bc-1-2c^2-bc+c^2}{bc-1-2c^2}=\frac{x(bc-1-2c^2+bc-b^2)}{b(bc-1-2c^2)}$$$$\frac{-(c^2+1)}{bc-1-2c^2}=\frac{x(2bc-1-2c^2-b^2)}{b(bc-1-2c^2}$$$$x=\frac{-b(c^2+1)}{2bc-1-2c^2-b^2}$$$$x=\frac{b(c^2+1)}{2c^2+b^2+1-2bc}$$And, $$y=1-\frac{x}{b}$$$$y=1-\frac{b(c^2+1)}{b(2c^2+b^2+1-2bc)}$$$$y=\frac{2c^2+b^2+1-2bc-c^2-1}{2c^2+b^2+1-2bc}$$$$y=\frac{(b-c)^2}{2c^2+b^2+1-2bc}$$Hence, $$K \equiv \left( \frac{b(c^2+1)}{2c^2+b^2+1-2bc} , \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)$$\par \textbf{\underline{Coodinate for Z}}: Note, that, since, $DD^*$ is the diameter of $\omega$, hence, $ZD^* \perp ZD$ or $ZD^* \perp AD$, therefore, $$m_{ZD^*} \cdot m_{AD}=-1$$Now we have eariler found out that, $$AD: y=\frac{(x-c)(c-b)}{bc-1-2c^2}$$$$AD:y=\frac{x(c-b)}{bc-1-2c^2} -\frac{c(b-c)}{bc-1-2c^2}$$We know that equation of a line can be represented in the form of $y=mx+b$, where $m$ is the slope of the line and $b$ is the $y-$intercept, hence using this we get, $$m_{AD}=\frac{(c-b)}{bc-1-2c^2}$$ Hence, $$ m_{ZD^*}=\frac{-1}{m_{AD}}=\frac{-1}{\frac{c-b}{bc-1-2c^2}}$$ $$m_{ZD^*}=\frac{2c^2+1-bc}{c-b}$$Let $Z \equiv (x',y')$ and recall that $D^* \equiv (b,bc+1)$, hence ,we can find equation of $ZD^*$, $$ZD^*: \frac{y-y'}{bc+1-y'}=\frac{x-x'}{b-x'}$$$$ZD^*:y=y'+\left( \frac{x-x'}{b-x'} \right) (bc+1-y')$$Using the earier we stated that, any line can be written in the form of $y=mx+b$, so using this, $$m_{ZD^*}=\frac{bc+1-y'}{b-x'}$$but, we eariler found out that, $$m_{ZD^*}=\frac{2c^2+1-bc}{c-b}$$$$\Longrightarrow \frac{2c^2+1-bc}{c-b}=\frac{bc+1-y'}{b-x'} ----(3)$$And, Since, $Z \in AD$, $$AD: y=\frac{(x-c)(b-c)}{bc-1-2c^2}$$$$y'=\frac{(x-c)(b-c)}{bc-1-2c^2}----(4)$$Solving $(3)$ and $(4)$ should yield the coordinates for $Z$, $$(3) \longrightarrow$$$$\frac{2c^2+1-bc}{c-b}=\frac{bc+1-y'}{b-x'}$$$$(2c^2+1-bc)(b-x')=(bc+1)(c-b)-y'(c-b)$$$$y'=\frac{(2c^2+1-bc)(b-x')+(bc+1)(b-c)}{b-c} -----(5)$$ Therefore, dividing $(4)$ and $(5)$, we have, $$\frac{y'}{y'}=\frac{\frac{(x'-c)(c-b)}{bc-1-2c^2}}{\frac{(2c^2+1-bc)(b-x')+(bc+1)(b-c)}{b-c}}$$$$\frac{bc-1-2c^2}{b-c}=\frac{(x'-c)(c-b)}{(2c^2+1-bc)(b-x')+(bc+1)(b-c)}$$$$\frac{bc-1-2c^2}{b-c}=\frac{x'(c-b) +bc-c^2}{2bc^2+b-b^2c-x'(2c^2+1-bc)+b^2c+b-bc^2-c}$$$$\frac{bc-1-2c^2}{b-c}=\frac{x'(c-b) +bc-c^2}{bc^2+2b-c-x'(2c^2+1-bc)}$$$$(bc-1-2c^2)(bc^2+2b-c-x'(2c^2+1-bc))=x'(c-b)(b-c)+(bc-c^2)(b-c)$$$$(bc-1-2c^2)(bc^2+2b-c)+x'(bc-1-2c^2)^2=c(b-c)^2-x'(b-c)^2$$$$b^2c^3+2b^2c-bc^2-bc^2-2b+c-2bc^4-4bc^2+2c^3+x'(bc-1-2c^2)^2=b^2c-2bc^2+c^3-x'(b-c)^2$$$$b^2c^3+2b^2c-bc^2-bc^2-2b+c-2bc^4-4bc^2+2c^3-b^2c+2bc^2-c^3=-x'((b-c)^2+(bc-1-2c^2))$$$$(b^2c-2bc^2-2b+c)(c^2+1)=-x'(b^2-2bc+c^2+b^2c^2+1+4c^4-2bc+4c^2-4bc^3)$$$$(b^2c-2bc^2-2b+c)(c^2+1)=-x'((c^2+1)(b^2-4bc+4c^2+1))$$$$x' \equiv - \left( \frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} \right)$$ And, Since, as above stated, $$y'=\frac{(x'-c)(c-b)}{bc-1-2c^2}$$$$y'=\frac{(-(\frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1})-c)(c-b)}{bc-1-2c^2}$$$$y'=\frac{-(\frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} +c)(c-b)}{bc-1-2c^2}$$$$y'=\frac{(\frac{b^2c-2bc^2-2b+c+b^2c-4bc^2+4c^3+c}{b^2-4bc+4c^2+1})(b-c)}{bc-1-2c^2}$$$$y'=\frac{2(b-c)(bc-1-2c^2)(b-c)}{(bc-1-2c^2)(b^2-4bc+4c^2+1)}$$$$y' \equiv \frac{2(b-c)^2}{b^2-4bc+4c^2+1}$$Hence, Combining we have the coordinates for $Z$ $$Z \equiv \left( - \left( \frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} \right) , \left( \frac{2(b-c)^2}{b^2-4bc+4c^2+1} \right) \right)$$ \par \textbf{\underline{Equation for line $ZD^*$ }}: Recall, $D^* \equiv (b,bc+1)$ and we have found out the coordinate for $Z$ above, Hence, $$ZD^* : \frac{y-(bc+1)}{\frac{2(b-c)^2}{b^2-4bc+4c^2+1} -(bc+1)}=\frac{x-b}{- \left( \frac{b^2c-2bc^2-2b+c}{b^2-4bc+4c^2+1} \right) -b}$$$$ZD^*: \frac{y-(bc+1)}{2(b-c)^2-(bc+1)(b^2-4bc+4c^2+1)}=\frac{x-b}{-(b^2c-2bc^2-2b+c+b^3-4b^2c+4bc^2+b)}$$$$ZD^*: \frac{y-(bc+1)}{-(b^2-2bc-1)(bc-2c^2-1)}=\frac{x-b}{-(b-c)(b^2-2bc-1)}$$ $$ZD^*: y=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$ \par \textbf{\underline{Equation for line AX}}: Recall, $A \equiv \left( \frac{bc-1}{2c} , \frac{c-b}{2c} \right)$ and $X \equiv \left( 0, \frac{bc-c^2}{bc+1} \right) $, Hence, $$AX: \frac{y- \frac{c-b}{2c}}{\frac{bc-c^2}{bc+1}-(\frac{c-b}{2c})}=\frac{x-\frac{bc-1}{2c}}{0-(\frac{bc-1}{2c})}$$$$AX: \frac{y-\frac{c-b}{2c}}{\frac{(bc-c^2)(2c)-(c-b)(bc+1)}{(bc+1)(2c)}} =\frac{x-(\frac{bc-1}{2c})}{-(\frac{bc-1}{2c})}$$$$AX: \frac{y-(\frac{c-b}{2c})}{\frac{b^2c+bc^2+b-2c^3-c}{bc+1}}=\frac{x-(\frac{bc-1}{2c})}{1-bc}$$$$AX: y-\left(\frac{c-b}{2c} \right)=\left( \frac{2cx-(bc-1)}{2c(1-bc)} \right) \left( \frac{b^2c+bc^2+b-2c^3-c)}{bc+1} \right)$$ $$AX: y=\frac{c-b}{2c}- \left( \frac{(2cx-(bc-1))(b^2c+bc^2+b-2c^3-c)}{2c(b^2c^2-1)} \right)$$ \par \textbf{\underline{Coordinate for L}}: Recall, that, $AX \cap ZD^*=L$ We have already found the equation for line $AX$ and line $ZD^*$, so now, we can find out the coordinates for $L$, $$AX: y=\frac{c-b}{2c}- \left( \frac{(2cx-(bc-1))(b^2c+bc^2+b-2c^3-c)}{2c(b^2c^2-1)} \right)$$$$ZD^*: y=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$Equating $y$, we get, $$\Longrightarrow \frac{c-b}{2c}- \left( \frac{(2cx-(bc-1))(b^2c+bc^2+b-2c^3-c)}{2c(b^2c^2-1)} \right)=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$$$\Longrightarrow -(b-c)^2(b^2c^2-1)-(2cx+1-bc)(b^2c+bc^2+b-2c^3-c)(b-c)$$$$=2c(bc+1)(b-c)(b^2c^2-1)+(2c)(bc-1-2c^2)(x-b)(b^2c^2-1)$$ $$\Longrightarrow -2b^3c^4x+4b^2c^5x-6c^3x-2cx-2b^3c^2x-2b^2cx+6bc^4x+6bc^2x-4c^5x+2b^2c^3x$$$$=2b^3c^3+2c^2+2b^3c^5-4bc+2b^2c^4+2b^2c^2-2bc^5-6bc^3+2c^4$$ $$\Longrightarrow -2c(c^2+1)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)x=2c(c^2+1)(bc-1)(b^2c+2b-c)$$$$\Longrightarrow x=\frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}$$ Now, we can use the value of $x$ to find out the value of $y$, $$y=(bc+1)+\left(\frac{bc-2c^2-1}{b-c} \right) (x-b)$$$$y=\frac{(bc+1)(b-c) + (bc-2c^2-1)(x-b)}{b-c}$$$$y=\frac{(b^2-bc^2+b-c)+(bc-2c^2-1) \left(\frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}-b \right)}{b-c}$$$$y=\frac{(bc+1)(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)-(bc-2c^2-1)(b^4c-b^3c^2+b^3-2b^2c+bc^2-b+c)}{(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}$$$$y=\frac{2b^4c-4b^3c^2+2b^3+2b^2c^3-4b^2c+2bc^2}{(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}$$$$y=\frac{2b(b-c)^2(bc+1)}{(b-c)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}$$$$y=\frac{2b(b-c)(bc+1)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}$$ \par \par $$L \equiv \left( \frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} , \frac{2b(b-c)(bc+1)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} \right)$$ \par \begin{flushleft} Now we'll calculate the equation for $YK$ \end{flushleft} \par \textbf{\underline{Equation for line YK}}: Recall, $Y \equiv \left( \frac{2bc+b^2c^2-c^2}{b(c^2+1)} , \frac{-c(b-c)^2}{b(c^2+1)} \right)$ and \par $K \equiv \left( \frac{b(c^2+1)}{2c^2+b^2+1-2bc} , \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)$ Hence, we have, $$YK : \frac{y-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{-c(b-c)^2}{b(c^2+1)}-\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}=\frac{x-\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{2bc+b^2c^2-c^2}{b(c^2+1)}-\left(\frac{b(c^2+1)}{2c^2+b^2+1-2bc}\right)} $$$$YK : \frac{y-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{c(b-c)^2}{b(c^2+1)}+\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}=\frac{x-\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{b(c^2+1)}{2c^2+b^2+1-2bc}-\left(\frac{2bc+b^2c^2-c^2}{b(c^2+1)} \right)} $$ Now if the coordinates of $L$ satisfy equation of $YK$, then we have indeed proved that $L$ lies on $YK$, so we'll do it by first computing the RHS and then LHS and then if RHS equals LHS, we're done with our claim, So here we go!, \par \textbf{\underline{LHS}}:\par $$\Longrightarrow \frac{y-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{c(b-c)^2}{b(c^2+1)}+\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}$$$$\Longrightarrow \frac{\left( \frac{2b(b-c)(bc+1)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} \right)-\left( \frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}{\frac{c(b-c)^2}{b(c^2+1)}+\left(\frac{(b-c)^2}{2c^2+b^2+1-2bc} \right)}$$$$\Longrightarrow \frac{\frac{(b^2-2bc+2c^2+1)(2b^2-2bc)(bc+1)-(b-c)^2(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}{(b^3c+b^2-2b^2c^2-3bc+2c^2+1)(b^2-2bc+2c^2+1)}}{\frac{c(b-c)^2(b^2-2bc+2c^2+1)+(b-c)^2(bc^2+b)}{(bc^2+b)(b^2-2bc+2c^2+1)}}$$$$\Longrightarrow \frac{\frac{b^4c-b^3c^2+b^3+2b^2c^3+2b^2c-bc^2+b+2c^3+c}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}}{\frac{(b-c)(b^2c-bc^2+b+2c^3+c)}{bc^2+b}}$$$$\Longrightarrow \frac{\frac{b^4c-b^3c^2+b^3+2b^2c^3+2b^2c-bc^2+b+2c^3+c}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}}{\frac{b^3c-2b^2c^2+b^2+3bc^3-2c^4-c^2}{bc^2+b}} ------(6)$$ \par \textbf{\underline{RHS}}: \par $$\Longrightarrow \frac{x-\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{b(c^2+1)}{2c^2+b^2+1-2bc}-\left(\frac{2bc+b^2c^2-c^2}{b(c^2+1)} \right)}$$$$\Longrightarrow \frac{\left( \frac{-(bc-1)(b^2c+2b-c)}{b^3c+b^2-2b^2c^2-3bc+2c^2+1} \right) -\frac{b(c^2+1)}{2c^2+b^2+1-2bc}}{\frac{b(c^2+1)}{2c^2+b^2+1-2bc}-\left(\frac{2bc+b^2c^2-c^2}{b(c^2+1)} \right)}$$$$\Longrightarrow \frac{- \left( \frac{(bc-1)(b^2+2b-c)(b^2-2bc+2c^2+1)+(bc^2+b)(b^3c+b^2-2b^2c^2-3bc+2c^2+1)}{(b^3c+b^2-2b^2c^2-3bc+2c^2+1)(b^2-2bc+2c^2+1) } \right) }{\frac{(bc^2+b)^2-(2bc+b^2c^2-c^2)(b^2-2bc+2c^2+1)}{(bc^2+b)(b^2-2bc+2c^2+1)}}$$ $$\Longrightarrow \frac{\frac{-b^5c^2+b^4c^3-2b^4c+3b^3c^2+b^3-b^2c^3-3b^2c+4bc^2+b-2c^3-c}{(b^3c+b^2-2b^2c^2-3bc+2c^2+1)(b^2-2bc+2c^2+1)}}{\frac{-b^4c^2+2b^3c^3-2b^3c-b^2c^4+6b^2c^2+b^2-6bc^3-2bc+2c^4+c^2}{(bc^2+b)(b^2-2bc+2c^2+1)}}$$$$\Longrightarrow \frac{\frac{-b^5c^2+b^4c^3-2b^4c+3b^3c^2+b^3-b^2c^3-3b^2c+4bc^2+b-2c^3-c}{b^3c+b^2-2b^2c^2-3bc+2c^2+1}}{\frac{-b^4c^2+2b^3c^3-2b^3c-b^2c^4+6b^2c^2+b^2-6bc^3-2bc+2c^4+c^2}{bc^2+b}} -----(7)$$\par Now note that, $$\left(-b^5c^2+b^4c^3-2b^4c+3b^3c^2+b^3-b^2c^3-3b^2c+4bc^2+b-2c^3-c \right) \cdot \left(b^3c-2b^2c^2+b^2+3bc^3-2c^4-c^2 \right) $$$$= -b^8c^3+3b^7c^4-3b^7c^2-5b^6c^5+8b^6c^3-b^6c+5b^5c^6-12b^5c^4-2b^5c^2+b^5-2b^4c^7+14b^4c^5+14b^4c^3-2b^4c$$$$-9b^3c^6-24b^3c^4+b^3+2b^2c^7+23b^2c^5+6b^2c^3-b^2c-14bc^6-9bc^4-bc^2+4c^7+4c^5+c^3$$$$=\left( b^4c-b^3c^2+b^3+2b^2c^3+2b^2c-bc^2+b+2c^3+c \right)$$$$ \cdot \left( -b^4c^2+2b^3c^3-2b^3c-b^2c^4+6b^2c^2+b^2-6bc^3-2bc+2c^4+c^2 \right)$$ So using this, ofcourse, $(6)$ and $(7)$ are equal, And Hence, $L$ lies on $KY$ Hence, our Claim is proved!! \par \par \begin{flushleft}

Draw $\ell$ parallel to $XC$ through $Y$, then $CY \perp \ell$ and let $\ell \cap AC=P$ and $\ell \cap BC=Q$, then, $\Delta XAC \sim \Delta YAP$ and $\Delta XBC \sim \Delta YBQ$ gives us, $$\frac{AX}{XC}=\frac{AY}{PY} -----(i)$$And, $$\frac{BX}{XC}=\frac{BY}{YQ} -----(ii)$$and, $$-1=(X,Y;A,B) \Longrightarrow \frac{AX}{AY}=\frac{BX}{BY} ------(iii)$$Hence, using $(i), (ii), (iii)$, it's easy to get, $$YP=YQ \Longrightarrow \angle ACY=\angle BCA$$Hence, $CY$ is the angle bisector of $\angle ACB$ Hence we're completed with the proof of our Lemma!!

$\angle YXE= 90^{\circ}-\angle FDX= 90^{\circ}- \angle FED= \angle YFE$ so $XFEY$ is cyclic.