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 $$PY=PB$$![enter image description here]( Idea: since $YA,YC$ are tangents to $\odot PAC$ we see that $PY$ is is the symmedian of $PAC$.then I can't it

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. ![enter image description here](

$PY=P'C$ by rotating around $A$, then $P'C=PB$ since the parallelograms $PABC$ and $PCC'P'$ are congruent.