14. As shown in Figure 1, given that the circumradius $R$ of the acute triangle $\triangle ABC$ is $1$, $\angle BAC = 60^{\circ}$, and the orthocenter and circumcenter of $\triangle ABC$ are $H$ and $O$ respectively. The line segment $OH$ intersects the extension of $BC$ at point $P$. Find:
(1) The area of the concave quadrilateral $ABHC$;
(2) The value of $\mathrm{PO} \cdot \mathrm{OH}$.
14. (1) As shown in Figure 2, connect $A H$, and 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 property of the Euler line, $A H=2 O D=1$.
By the Law of Sines, $B C=2 R \sin A=\sqrt{3}$, so 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 $H$ is the orthocenter of $\triangle A B C$,
$$
\angle B H C=180^{\circ}-\angle B A C=120^{\circ}=\angle B O C \text {. }
$$

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

Therefore, $P O^{2}-P O \cdot P H=1$, which means $P O \cdot O H=1$.