show that $PY=PB$ is such $\Delta YAC$ is equilateral If P is a point on $B$ median of $ABC$ with $\angle B=60$,Let $X$ be the reflection $B$ wrt midpt of $AC$,and $X=P$,Let $Y$ be a point on the same side of $AC$ as $B$ such that $YAC$ is equilateral,show that
$$PY=PB$$
Idea: since $YA,YC$ are tangents to $\odot PAC$ we see that $PY$ is is the symmedian of $PAC$.then I can't it
 A: With a couple of additional points is pretty trivial. $P'$ and $C'$ are chosen in such a way that both $CBC'$ and $PAP'$ are equilateral.

$PY=P'C$ by rotating around $A$, then $P'C=PB$ since the parallelograms $PABC$ and $PCC'P'$ are congruent.
A: Let $M$ be the middle point of $AC$ and $\omega$ be the reflection of the circumcircle $ABC$ with respect to $M$. Since $P \in \omega$ and $PB = 2 PM$, your problem is restated as: 

For every $P \in \omega$, $PY = 2 PM$. 

The circles of Apollonius theorem states that points $P$ satisfying $PY = 2 PM$ form some circle $\omega'$. This equality is easily verified when $P$ coincides with $A$, $C$ or the point $Y'$ symmetric to $Y$ with respect to $M$. This proves that $A,C,Y' \in \omega'$ and hence $\omega'$ coincides with $\omega$. 

I'm using the following theorem: 

Let $X_1,X_2$ be distinct points and $\alpha > 1$. Then the locus of points $P$ such that $PX_1 = \alpha PX_2$ is a circle. 

Its proof can be found here. In the above, $X_1 = Y$, $X_2 = M$, $\alpha = 2$. 
This is not the easiest way (as shown by Jack D'Aurizio), but I think this point of view is also nice.  
A: Constructions and notations. Let $M$ be the midpoint of segment $AC$. Since, by construction, point $P$ is the symmetric image of point $B$ with respect to $M$, quad $ABCP$ is a parallelogram, so $M$ is the midpoint of $PB$ as well. Let $N$ be the midpoint of segment $BY$. Then $MN$ is a midsegment in triangle $BPY$ so $MN$ is parallel to $PY$ and $MN = \frac{1}{2} PY$. Furthermore, $MB = \frac{1}{2} PB$ so if I prove that $MN = MB$ then immediately $PY = PB$.

Claim. $\,\,\, MN = MB$
Proof: Let $L$ be the midpoint of $AY$. Then $ML$ is a midsegment of equilateral triangle $ACY$, so triangle $ALM$ is also equilateral and therefore $$ML = MA = LA = LY = \frac{1}{2} AC$$ Furthermore, let $H$ be the orthogonal projection of point $A$ on $BC$, so $\angle \, AHC = \angle \, AHB = 90^{\circ}$. Since $M$ is the midpoint of $AC$
$$MH = MA = MC = \frac{1}{2} AC$$ which means that 
$$ML = \frac{1}{2} AC =  MH \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (1)$$
Moreover, since $AHB$ is right-angled triangle with angle $\angle \, ABH = \angle \, ABC = 60^{\circ}$
$$HB = \frac{1}{2} AB$$ However, since $L$ and $N$ are midpoints of segments $AY$ and $BY$, the segment $LN$ is a midsegment of triangle $ABY$ so it is parallel to $AB$ and half of it in length. Thus
$$LN = \frac{1}{2} AB = HB \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (2)$$
Finally, since $\angle \, ABC = \angle \, AYC = 60^{\circ}$ the quad $ACBY$ is cyclic, so after some angle chasing one can show that $$\angle \, NLM = \angle \, BHM \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (3)$$
By (1), (2) and (3) we conclude that triangles $LMN$ and $HMB$ are congruent and therefore $$MN = MB$$
