In this scale drawing, A, B and C are three corners of a park ABCD with C due north of B.

1. Construct the perpendicular bisector of $BC$ accurately.

Set the compass radius to more than half of $BC$. With centre $B$, draw arcs on both sides of $BC$. Using the same radius with centre $C$, draw arcs to intersect the first pair of arcs. Join the two arc intersection points with a straight line.

This line is the perpendicular bisector of $BC$.

1. Draw the locus satisfying the bearing condition from $A$.

From $A$, draw a north line. Measure a clockwise angle of $065^\circ$ from north and draw the ray from $A$ on this bearing.

2. Use the equidistant condition to locate and label $D$.

Since $D$ is equidistant from $B$ and $C$, $D$ lies on the perpendicular bisector of $BC$ constructed in part (a). Mark the intersection of the $065^\circ$ bearing ray from $A$ with this perpendicular bisector and label it $D$.

1. Construct the angle bisector of $\angle ABC$ accurately.

Using a compass, draw an arc centred at $B$ to cut $BA$ and $BC$. From the two intersection points, draw equal-radius arcs to meet inside $\angle ABC$. Draw a straight line from $B$ through the intersection of these arcs. This line is the bisector of $\angle ABC$.

1. Identify the two locus boundaries and choose the correct side of each boundary.

Points closer to line $BC$ than to line $AB$ lie on the $BC$ side of the angle bisector of $\angle ABC$.

Points closer to $B$ than to $C$ lie on the $B$ side of the perpendicular bisector of $BC$.

Therefore, shade the intersection of these two half-regions inside the park $ABCD$.