The diagram shows a triangle ABC. On the diagram,

1. Construct equal arcs from the vertex of the angle.

Place the compass point at $B$. Draw an arc cutting $BA$ and $BC$ at two points.

2. Locate a point inside the angle equidistant from the two arms.

Without changing the compass width, place the compass point on each of the two arc-intersection points and draw two arcs inside $\angle ABC$ so that the arcs intersect.

3. Draw the angle bisector.

Use a ruler to draw a straight line from $B$ through the intersection of the two arcs. This line is the bisector of $\angle ABC$.

1. Construct equal arcs from $B$ and $C$ to locate two points equidistant from both endpoints.

Set the compass to a radius greater than half of $BC$. With centre $B$, draw arcs above and below $BC$. Without changing the compass width, with centre $C$, draw arcs above and below $BC$ so that they intersect the first arcs.

2. Draw the perpendicular bisector through the two arc-intersection points.

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

1. Identify and label the common point satisfying both equidistance conditions.

Since $P$ is equidistant from $B$ and $C$, it lies on the perpendicular bisector of $BC$.
Since $P$ is equidistant from $AB$ and $BC$, it lies on the angle bisector of $\angle ABC$.
Therefore, label $P$ at the intersection of these two constructed lines.