jmc

Let $h$ be the distance from $P$ to $CB$. The area of triangle $ABC$ is $\frac{1}{2}(AC)(CB)$. The area of triangle $PBC$ is $\frac{1}{2}(h)(CB)$. Therefore, the area of triangle $PBC$ is less than one-half of the area of triangle $ABC$ if $h<AC/2$. This happens if $P$ falls below the dashed line whose endpoints are the midpoints $D$ and $E$ of $\overline{AC}$ and $\overline{AB}$. Triangle $ADE$ is similar to triangle $ACB$, so the ratio of the area of triangle $ADE$ to the area of triangle $ACB$ is ${\left(\frac{AD}{AC}\right)}^2=\frac{1}{4}$. Therefore, the ratio of the area of trapezoid $DEBC$ to the area of triangle $ABC$ is $1-\frac{1}{4}=\boxed{\frac{3}{4}}$.