Prove that the radical axis of $\omega_{B} '$ and $\omega_{C} '$ halves the perimeter of $ABC$ 
$\omega_{B}$ and $\omega_{C}$ are excircles of triangle $ABC$. The circle $\omega_{B} '$ is symmetric to $\omega_{B}$ with respect to the midpoint of $AC$, the circle $\omega_{C} '$ is symmetric to $\omega_{C}$ with respect to the midpoint of $AB$. Prove that the radical axis of $\omega_{B} '$ and $\omega_{C} '$ halves the perimeter of $ABC$

My Progress: Please refer the diagram below

here $I_c$ is the centre of  $\omega_{C}$ , $O$ is the centre of  $\omega_{C'}$,$I_B$ is the centre of  $\omega_{C'}$ and $W$ is the centre of $\omega_{B'}$
$Y$ and $X$ are midpoints of $AB$ and $AC$.
and $P,M,Q$ are touch points of $\omega_{C}$ to $BC,BA,AC$
and $S,X,R$ are touch points of $\omega_{B}$ to $BC,AC,AB$
and $N$ is the touch point of $omega_{C'}$ to $BA$
and $\omega_{B'}$, $\omega_{B'}$ intersect at $F$ and $J$
and $K=FJ\cap BC$ and $V$ is the centre of $A$-excircle .
Claim: $AB$ is tangent to $\omega_{C'}$
Proof: drop perpendicular  from $O$ to $AB$ and intersect at $N'$.
then we have Then $I_CMY$ and $ON'Y$ are similar, giving $MY=N'Y$ which proves $N=N'$
and hence , we have $AB$ is tangent to $\omega_{C'}$
Claim: $AC$ is tangent to $\omega_{B'}$
similar proof
Hence we have  $A,F,J$ collinear ( radical axis)
Now,  we need to show that $AB+BK= AC+CK$ or it is enough to show that $AB+BK= AM+MB+BK= AQ+BP+BK=AS+CK= CS+AR+CK$ , but we know $CP=BS$.  So we have $BP=CS$. Hence enough to show that $AQ+BK= AR+CK$.
So we have to prove that $K$ is  tangency of $A$ excircle to $BC$ , which I am not able to.

Now, I couldn't think of any synthetic approach but I think coordinates can help us .
So , I assigned $B=(0,0)$ , $C=(1,1)$ and $A=(a,b)$ . Let $s$ be the semi-perimetre of the triangle $ABC$.
So $P=(1-s,0)$ , $I_C= (1-s,s\tan{C/2})$ .
$S=(s,0)$, $I_B=(s,s\tan{B/2})$
$Y=(a/2,b/2)$
$X=({a+1}/2,{b+1}/2)$
$O= (1-s-a/2,s\tan{C/2} -b/2) $
$W= (s-{a+1}/2,s\tan{B/2}-{b+1}/2) $
then I took the equations of $\omega_{B} '$ and $\omega_{C}$ , and then I tried to find the radical axis and then tried to find coordinates of $K$. But we just to find the X - coordinates since $K $ lies in the x axis , but I gave up in the middle , since it was becoming to complex .
After finding the coordinates of $K$, I was thinking on finding coordinates of $V$
and then show that $VK\perp BC$.
Thanks in advance .
 A: Okay, so I realized that my initial proof was a fake solve  but it was fixable. BTW stalute to anyone who solved it without geogebra XD.
Anyways, here's the proof:

Notations. $I$ is incenter. $M_{P_1P_2}$ is midpoint of $P_1P_2$. $X$ and $Y$ are centers of $\omega_B'$ and $\omega_C'$ respectively. $\odot(I)\cap\{BC,CA,AB\}=\{D,E,F\}$ and $\omega_B\cap AC = E_B,\omega_C\cap AB = F_C,\omega_A\cap BC= D_A$. Let $I_{AC}$ be reflection of $I$ in $AC$.


Claim 1. $X-I-E$ are collinear.
Proof of claim 1. Note that as $\mathcal H_{M_{AC}}(-1): \odot(I_B)\mapsto \omega_B'\implies \mathcal H_{M_{AC}}(-1): E_{B}\mapsto E$ and as tangency is preserved in homothety, we get $\odot(I)$ and $\omega_B'$ are internally tangent at $E$.$\tag*{$\blacksquare$}$
Claim 2. $A,I_{AC},C, X$ are concyclic.
Proof of claim 2. Note that $\odot(AICI_B)$ and $\odot(AI_{AC}C)$ are reflection of each other about $AC$. Thus, $\mathcal H_{M_{AC}}(-1): \odot(AICI_B)\mapsto \odot(AI_{AC}C)$ as $M_{AC}$ is midpoint of common chord of $\odot(AIC)$ and $\odot(AI_{AC}C)$. Thus, as $\mathcal H_{M_{AC}}(-1): I_{B}\mapsto X$ and $I_B\in\odot(AIC)\implies X\in\odot(AI_{AC}C)$. $\tag*{$\blacksquare$}$
Claim 3. $M_{AB}M_{BC}$ is polar of $X$ with respect to $\odot(I)$.
Proof of claim 3. From Claim 1. adn Claim 2. we get $X=II_{AC}\cap \odot(AI_{AC}C)$. Now note that $XI\perp AC$ and $\mathcal H_{E}(-1): I\mapsto I_{AC}\in\odot(AXC)\implies I$ is orthocenter of $\triangle AXC$. Thus, pole of $XA$ wrt $\odot(I)$ lies on $CI$ as $XA\perp CI$. Also, as $EF$ is polar of $A$ wrt $\odot(I)$, by La Hire's theorem, we get $Z$ lies on polar of $X$ wrt $\odot(I)$ where $Z:=CI\cap EF$. Also, as polar of $X$ wrt $\odot(I)$ is parallel to $AC$, from Iran's lemma, we get $M_{AB}M_{BC}$ is the polar of $X$ wrt $\odot(I)$.$\tag*{$\blacksquare$}$
Claim 4. $M_{BC}I\perp XY$.
Proof of claim 4. From Claim 3. and symmetry, we get, $M_{AB}M_{BC}$ and $M_{AC}M_{BC}$ are polar of $X$ and $Y$ wrt $\odot(I)$ and thus, by La Hire's theorem, we get $XY$ is polar of $M_{BC}$ wrt to $\odot(I)$. $\tag*{$\blacksquare$}$
Its well-known that $M_{BC}I\| AD_A\overset{\textbf{Claim 4.}}{\implies} AD_A\perp XY$ and as $A\in\mathcal R(\omega_B',\omega_C')\implies AD_A=\mathcal R(\omega_B',\omega_C')$ and as concluded by you that it is enough to show $AD_A$ is radical axis, we are done!
