The diagram shows the map of a park in the shape of a quadrilateral ABCD. The scale of the map is 1 cm to 10 m.

1. Construct equal reference points on the two arms of the angle.

With centre $B$, draw an arc cutting $BA$ and $BC$ at two points, say $P$ and $Q$, respectively.

2. Locate a point equidistant from the two arms of the angle and draw the bisector.

Using the same compass radius, draw arcs from $P$ and $Q$ to meet inside $\angle ABC$ at $R$. Draw the straight line ray $BR$. This ray bisects $\angle ABC$.

1. Construct two equal-radius arc intersections from endpoints $A$ and $B$.

Set the compass radius to more than half of $AB$. With centre $A$, draw arcs above and below $AB$. Without changing the compass radius, with centre $B$, draw arcs cutting the first pair of arcs at two points.

2. Draw the perpendicular bisector of $AB$.

Use a ruler to join the two arc-intersection points. This line is the perpendicular bisector of $AB$.

1. Identify and shade the required intersection of the two locus regions.

The boundary for points closer to $AB$ than to $BC$ is the angle bisector of $\angle ABC$. Since the region must be closer to $AB$, choose the side of this bisector containing side $AB$.

The boundary for points closer to $B$ than to $A$ is the perpendicular bisector of $AB$. Since the region must be closer to $B$, choose the side of this perpendicular bisector containing $B$.

Shade the overlap of these two chosen regions inside $ABCD$.

1. Convert the real distance from $D$ to the map distance.

Scale: $1\text{ cm}$ represents $10\text{ m}$.

Therefore $30\text{ m}$ represents
$30 \div 10 = 3\text{ cm}.$

2. Construct the two required loci and mark their intersection inside the park.

Draw a circle with centre $D$ and radius $3\text{ cm}$.

Since $T$ is equidistant from $A$ and $B$, $T$ must lie on the perpendicular bisector of $AB$ constructed in part (b).

Mark the point where the circle intersects the perpendicular bisector of $AB$ inside $ABCD$, and label it $T$.