14 As shown in the figure, given that the circumradius $R=1$ of the acute triangle $\triangle ABC$, $\angle BAC=60^{\circ}$, and the orthocenter and circumcenter of $\triangle ABC$ are $H$ and $O$ respectively, the line segment $OH$ intersects the extension of $BC$ at point $P$.
(1) Find the area of the concave quadrilateral $ABHC$;
(2) Find the value of $PO \cdot OH$.
(14) (1) As shown in the figure, connect $A H$, draw $O D \perp B C$ at point $D$. Since $O$ is the circumcenter of $\triangle A B C$ and $\angle B A C=60^{\circ}$, we have
$$
\begin{array}{l}
\angle B O C=2 \angle B A C=120^{\circ}, \\
O D=O C \cos 60^{\circ}=\frac{1}{2} .
\end{array}
$$

By the properties of the Euler line, $A H=2 O D=1$. By the Law of Sines, we have
$$
B C=2 R \sin \angle B A C=\sqrt{3},
$$

Thus, the area of the concave quadrilateral $A B H C$ is $\frac{1}{2} A H \cdot B C=\frac{\sqrt{3}}{2}$.
(2) Since
$$
\angle B O C=2 \angle B A C=120^{\circ},
$$

and because $H$ is the orthocenter of $\triangle A B C$, we have
$$
\begin{aligned}
\angle B H C & =180^{\circ}-(\angle H B C+\angle H C B) \\
& =180^{\circ}-\angle B A C \\
& =120^{\circ} .
\end{aligned}
$$

Therefore, points $B, C, H, O$ are concyclic, and $P O \cdot P H=P B \cdot P C$.
Since $O$ is the circumcenter of $\triangle A B C$, and by the power of a point theorem, we have
$$
P B \cdot P C=P O^{2}-R^{2} \text {. }
$$

Thus,
$$
P O \cdot P H=P B \cdot P C=P O^{2}-R^{2},
$$

which implies
$$
P O^{2}-P O \cdot P H=1,
$$

or $\mathrm{PO} \cdot O H=1$.