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$P$ is inside $\Delta ABC$ such that $AB,BC,CA$ are tangents to the circumcircles of $\Delta BPC,\Delta CPA,\Delta APB$ respectively . Extended $AP,BP,CP$ cuts the circumcircle of $\Delta ABC$ at $A',B',C'$ .

Prove that power of $A$ (with respect to $\odot B'PC$ ),power of $B$ (wrt $\odot C'PA$ ),power of $C$ (wrt $\odot A'PB$ ) are equal .

(selfmade)

I might be wrong, but $A,B,C$ all lie on the circles they are meant to be respected to. So they have equal power $0$.

Nirjhor wrote: $A,B,C$ all lie on the circles they are meant to be respected to.

Not true.

First, prove that, $AB',$ etc. are tangent to $\odot B'PC,$ etc. Then, the powers of $A, B, $ and $C$ w.r.t. the given circles are equal to the squares of $AB', BC', CA'$. So, we'll be done if we can show $AB'=BC'=CA'$.

Now, for the proof.
$P$ is a Brocard Point of $\triangle ABC$. $\therefore \angle ABP=\angle BCP=\angle CAP$.
So,
$\angle AB'P=\angle AB'B=\angle ACB=\angle ACP+\angle PCB=$ $\angle ACP+\angle PBA=\angle ACP+\angle B'BA=\angle ACP+\angle B'CA=\angle B'CP$
$\therefore AB'$ is tangent to $\odot B'PC$. Similarly, $BC', CA'$ are tangent to $\odot C'PA, \odot A'BP$.
Finally, $AB', BC', CA'$ subtend equal angles on the circle. $\therefore AB'= BC'= CA'$, and, we're done.

Sorry I messed the construction up.