Construct triangle ABC where AC is 10 cm and angle BAC = 40°. AB has already been drawn. Construct the perpendicular bisector of AB, the angle bisector of angle BAC. Mark clearly a possible point which is inside triangle ABC, equidistant from A and B, and is nearer to AC than to AB. Label this point P.

1. Construct the ray from $A$ making $40^\circ$ with $AB$.

At $A$, use a protractor to draw a ray making an angle of $40^\circ$ with the given line $AB$ on the intended side of $AB$ for the triangle.

2. Locate $C$ so that $AC=10\text{ cm}$ and complete the triangle.

On the $40^\circ$ ray, mark point $C$ such that $AC=10\text{ cm}$. Join $C$ to $B$ to complete $\triangle ABC$.

1. Construct equal arcs from $A$ and $B$ to locate two points equidistant from $A$ and $B$.

Set the compass radius to more than half of $AB$. With centre $A$, draw arcs above and below $AB$. Using the same compass radius and centre $B$, draw arcs cutting the first pair of arcs at two points.

2. Draw the perpendicular bisector through the two arc intersections.

Join the two arc intersection points with a straight line. This line is the perpendicular bisector of $AB$.

1. Construct the locus points needed to bisect $\angle BAC$.

With centre $A$, draw an arc cutting $AB$ and $AC$ at two points. Using the same radius from these two points, draw two arcs intersecting inside $\angle BAC$. Draw the straight line from $A$ through this intersection point.

1. Identify and mark a valid position for $P$ using the constructed loci.

Mark $P$ inside triangle $ABC$ such that:

• $P$ lies on the perpendicular bisector of $AB$, so $PA=PB$,
• $P$ lies on the side of the angle bisector of $\angle BAC$ nearer to $AC$, so $P$ is nearer to $AC$ than to $AB$.

Hence $P$ satisfies both conditions.