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

1. Construct equal-radius arcs from $A$ and $B$ and draw the bisector through their intersections.

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 to intersect the first pair of arcs. Join the two arc intersection points with a straight line. This line is the perpendicular bisector of $AB$.

1. Construct the angle bisector of $\angle ABC$ using equal-radius arcs.

With centre $B$, draw an arc cutting $BA$ and $BC$ at two points. Using the same radius from these two points, draw arcs that intersect inside $\angle ABC$. Join $B$ to this intersection point with a ruler. This line is the angle bisector of $\angle ABC$.

1. Convert the actual distance from $D$ to a map distance.

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

$55\text{ m}\div 10=5.5\text{ cm}$

So $S$ must be $5.5\text{ cm}$ from $D$ on the map.

2. Locate the point satisfying both conditions.

Since $S$ is equidistant from $A$ and $B$, $S$ lies on the perpendicular bisector of $AB$.

Using compasses, draw an arc or circle with centre $D$ and radius $5.5\text{ cm}$.

The required point $S$ is the intersection of this circle with the perpendicular bisector of $AB$ that lies inside quadrilateral $ABCD$.

1. Identify and label the intersection satisfying both equidistance conditions.

Since $T$ is equidistant from $AB$ and $BC$, it lies on the angle bisector of $\angle ABC$.

Since $T$ is equidistant from $A$ and $B$, it lies on the perpendicular bisector of $AB$.

Therefore, label
$T=\text{ intersection of the angle bisector of }\angle ABC\text{ and the perpendicular bisector of }AB$
inside the quadrilateral $ABCD$.