Triangle - coordinate geometry problem Let ABC be a triangle. Let BE and CF be internal angle bisectors of B and C
respectively with E on AC and F on AB. Suppose X is a point on the segment CF
such that AX is perpendicular to CF; and Y is a point on the segment BE such that AY perpendicular BE. Prove
that XY = $\frac{b+c-a}{2}$, where BC = a , CA = b, AB = c.
Please give a proof using coordinate geometry.
My solution:
I took a general triangle with two vertices lying on x-axis and one vertex lying on y-axis.
Found the equation of angle bisectors of B and C.
Found the foot of perpendicular from A on CF and BE as X and Y respectively.
Found XY but the answer was not matching . Please help.
Again I request, please give a proof using coordinate geometry only.  
 A: The following is a diagram. And $B$ is the origin and assume each point $P$ has coordinate $(p_1,p_2)$.



*

*Point $A,B,C$



From the length three edges. We can immediately write $B(0,0)$ and $C(a,0)$. Then solve $\displaystyle A(\frac{a^2-b^2+c^2}{2a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{2a})$ from $a_1^2+a_2^2=c^2,~(a_1-a)^2+a_2^2=b^2$.



*

*Line $BE,CF$



Noting that 
  $$\displaystyle k_{BE}=\tan\frac B2=\frac{\tan B}{1+\sqrt{1+\tan^2B}}=\frac{k_{AB}}{1+\sqrt{1+k_{AB}^2}}=\frac{a_2}{a_1+\sqrt{a_1^2+a_2^2}}=\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{(a+c)^2-b^2}$$
  $$\displaystyle k_{CF}=\frac{-a_2}{(a-a_1)+\sqrt{(a-a_1)^2+(-a_2)^2}}=\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{(a+b)^2-c^2}$$
The equation of $BE$ is $k_{BE}x-y=0$ and $CF$, considering the line passes point $C$, $k_{CF}(x-a)-y=0$



*

*Point $X,Y$



Let $X$ be $\displaystyle (x_1,k_{CF}(x_1-a))$, $Y$ be $\displaystyle (y_1,k_{BE}y_1)$ since they are on lines $CF$, $BE$ respectively. According to orthogonal relation, we have
$\displaystyle AX\perp CF\Longleftrightarrow k_{AX}k_{CF}=\frac{a_2-k_{CF}(x_1-a)}{a_1-x_1}k_{CF}=-1\\
\displaystyle AY\perp BE\Longleftrightarrow k_{AY}k_{BE}=\frac{a_2-k_{BE}y_1}{a_1-y_1}k_{BE}=-1$
We obtain $\displaystyle X(\frac{3a^2-2ab-b^2+c^2}{4a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{4a}),Y(\frac{a^2-b^2+2ac+c^2}{4a},\frac{\sqrt{(b+c-a)(c+a-b)(a+b-c)(a+b+c)}}{4a})$ and hence the distance
  $$d(X,Y)=y_1-x_1=\frac {b+c-a}2$$

A: $A=(\frac{a^2-b^2+c^2}{2·a},\frac{2·S}{a}) \;  \; S=ΔABC$
$B=(0,0)$
$C=(a,0)$
Let G be a point such that the triangle ABG is isosceles.
$ΔABG, AB=BG, BX⟂AG$
Applicating section formula
$r_G=\frac{c}{a-c}$
$G=(\frac{xB+r_G·xC}{1+r_G},\frac{yB+r_G·yC}{1+r_G})$
$G=(c,0)$
X is midpoint of the segment AG
$X=(\frac{xC+xG}{2},\frac{yC+yG}{2})$
$X=(\frac{(a+b+c)·(a-b+c)}{4·a},\frac{S}{a})$
Let H be a point such that the triangle ACH is isosceles.
$ΔACH, AC=CH, CY⟂AH$
$r_H=\frac{a-b}{b}$
$H=(\frac{xB+r_H·xC}{1+r_H},\frac{yB+r_H·yC}{1+r_H})$
$H=(a-b,0)$
Y is midpoint of the segment AH
$Y=(\frac{xA+xH}{2},\frac{yA+yH}{2})$
$Y=(\frac{3·a^2-b^2+c^2-2·a·b}{4·a},\frac{S}{a})$
$XY=\sqrt{(xX-xY)^2+(yX-yY)^2}=\sqrt{(b+c-a)^2+0}=b+c-a$

