Proof that line $HG$ bisects the perimeter of $ABC$ 
Question. From this picture, $D$ and $E$ are excenters of $ABC$, and $G$ and $H$ are midpoints of $AB$ and $KL$. 


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*Prove that $HG$ bisects the perimeter of $ABC$. In other words, prove that $AX=CX+CB$.

*Prove that $HG$ is parallel to the angle bisector of angle $C$.
My Approach. I guess that $CX$=$CY$. If so, then we can prove problem 2) easily. And, we can change problem 1) as $AX=BY$.
 A: Here is an approach.
1)Show that $AK=BL$ (this is easy)
2) From 1) note that $G$ will have the same power with respect to both circles.
3) Prove that $H$ has the same power with respect to both circles (this is the difficult part)
Hint: Extend $KL$, so that $K'$ is the intersection with the circle on the left and $L'$ is the intersection with the circle on the right. Note that the power of K w.r.t the circle on the right is the same as the power of L w.r.t the circle on the left, from this conclude that $KK'=LL'$ and therefore the power of $H$ w.r.t both circles is the same.
4) Since the locus of the points that have the same power with respect to both circles is a line, then we have that this line is $GH$.
5) Since $X$ is in the line $GH$, then the power of $X$ with respect to both circles are the same. From this we get $KX=CX+CL$ and therefore, $AX=CX+CB$. This proves the first part.
6) Since $Y$ is in the line $GH$, then the power of $Y$ with respect to both circles are the same. From here, we get $CK-CY=CY+CL$. Now, using that  $KX=CX+CL$ (by 5) conclude that $CX=CY$.
