One, (50 points) In $\triangle ABC$, $AB=AC$. Try to find a necessary and sufficient condition that $\triangle ABC$ must satisfy so that there exists a point $P$ inside $\triangle ABC$ that meets the following conditions:
(1) $PB=AB$;
(2) $\angle ABP=2 \angle ACP$.

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As shown in Figure 2, let the circles with radius $AB$ and centers at $A$ and $B$ intersect at another point $Q$. Clearly, $\triangle QAB$ is an equilateral triangle, and $AQ = AC, \angle AQC = \angle ACQ$.

Since point $P$ is inside $\triangle ABC$ and $PB = AB$, point $P$ lies on the arc $\widehat{QQ}$. Also, $\angle AQP = \frac{1}{2} \angle ABP = \angle ACP$, thus,
$\angle PQC = \angle PCQ$.
Hence, $PQ = PC$.
Let the arc $\overparen{AQ}$ intersect $BC$ at point $K$. From the above analysis, we know:
A point $P$ exists inside $\triangle ABC$ such that $PB = AB$ and $\angle ABP = 2 \angle ACP$ if and only if the perpendicular bisector of segment $QC$ intersects the interior of arc $\overparen{AK}$ (excluding points $A$ and $K$), which is equivalent to the angle bisector $l$ of $\angle QAC$ intersecting the interior of arc $\overparen{AK}$ at a point.
Consider two extreme cases:
(i) $AC$ coincides with $AQ$. In this case, $\angle BAC = 60^\circ$.
(ii) The angle bisector $l$ is tangent to $\odot B$ at point $A$. In this case, by $l \perp AB$ and $\angle BAQ = 60^\circ$, the angle between $AQ$ and $l$ is $90^\circ - 60^\circ = 30^\circ$.
Thus, $\angle QAC = 2 \times 30^\circ = 60^\circ$, $\angle BAC = \angle BAQ + \angle QAC = 120^\circ$.
From the above sufficient and necessary condition, we have:
Point $P$ exists
$\Leftrightarrow$ the angle bisector $l$ of $\angle QAC$ intersects the interior of $\overparen{AK}$ at a point $\Leftrightarrow 60^\circ < \angle BAC < 120^\circ$ (otherwise, there would be no intersection).
Therefore, the required sufficient and necessary condition is $60^\circ < \angle A < 120^\circ$.
Note: We can prove that $\angle PCB = 30^\circ$.
Let $\angle CAP = \angle QAP = \alpha$, then
\[
\begin{array}{l}
\angle QAC = 2\alpha, \angle PBQ = 2 \angle QAP = 2\alpha, \\
\angle CBQ = \frac{1}{2} \angle QAC = \alpha.
\end{array}
\]

So, $\angle PBC = \alpha = \angle QBC$.
Thus, $BC$ is the perpendicular bisector of segment $PQ$.
Let $\angle PCB = \angle QCB = \gamma$, and $l$ intersects $QC$ at point $D$.
By $PD \perp QC$ and $CB \perp PQ$, we have $\angle QPD = \angle QCB = \gamma$.
Also, $\angle QPD = \frac{1}{2} \angle ABQ = \frac{1}{2} \times 60^\circ = 30^\circ$, hence $\angle PCB = \gamma = \angle QPD = 30^\circ$.
Note: This problem can also be solved using the "Fermat point" or "trigonometric methods".