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$P$ is the point where $\overline{BY}$ and $\overline{CZ}$ cross. $\Delta ABC$ is isosceles, and proving that $\Delta BPC$ is isosceles will be enough to show that $\overline{PY}=\overline{PZ}$, but I cannot determine a way to show that $\Delta BPC$ is isosceles.
This is not true in general.
There can be two points $Y_1,Y_2$ on the side $AC$ such that $BY_1=BY_2$ where $Y_1$ is nearer to $A$.
On the other hand, there can be two points $Z_1,Z_2$ on the side $AB$ such that $CZ_1=CZ_2$ where $Z_1$ is nearer to $A$.
Now choose $Y_1,Z_2$. We can have $BY_1=CZ_2$, but $PY_1=PZ_2$ does not always hold.