Construction of ABC
Given:
- AB = 5 cm
- ∠A = 40 degrees
- ∠B = 110 degrees

To construct triangle ABC, we can follow the given measurements and steps:

1. Draw a line segment AB of length 5 cm.
2. At point A, use a protractor to measure ∠A = 40 degrees. Mark the vertex of the angle.
3. From the vertex of ∠A, draw a ray extending towards the interior of the angle.
4. At point B, use a protractor to measure ∠B = 110 degrees. Mark the vertex of the angle.
5. From the vertex of ∠B, draw a ray extending towards the interior of the angle.
6. The intersection point of the two rays is point C, completing triangle ABC.

Locating the Circumcentre
To locate the circumcentre of triangle ABC, we need to find the point where the perpendicular bisectors of the sides of the triangle intersect. Let's find the perpendicular bisectors of sides AB, BC, and AC.

- Perpendicular Bisector of AB:
1. Find the midpoint M of side AB by measuring the segment and marking the midpoint.
2. Draw a line perpendicular to AB passing through M.
3. Repeat steps 1 and 2 for side BC and AC.

- Perpendicular Bisector of BC:
1. Find the midpoint N of side BC by measuring the segment and marking the midpoint.
2. Draw a line perpendicular to BC passing through N.

- Perpendicular Bisector of AC:
1. Find the midpoint O of side AC by measuring the segment and marking the midpoint.
2. Draw a line perpendicular to AC passing through O.

The point of intersection of the perpendicular bisectors is the circumcentre of triangle ABC. Label this point as O.

Measuring the Circumradius
The circumradius of a triangle is the distance between the circumcentre and any vertex of the triangle. To measure the circumradius, we need to find the length of OA, OB, or OC.

1. Measure the distance between the circumcentre O and any vertex, let's say vertex A. Label this distance as R.
2. The length R represents the circumradius of triangle ABC.

Related Theorem: One and Only Circle Passes Through Any Three Points in the Plane
The theorem states that given any three non-collinear points A, B, and C in the plane, there is one and only one circle that passes through all three points.

Proof:
1. Consider three non-collinear points A, B, and C.
2. Construct the perpendicular bisectors of the sides AB, BC, and AC as explained earlier.
3. The perpendicular bisectors intersect at a point O, which is the circumcentre of triangle ABC.
4. The distance between the circumcentre O and any vertex of the triangle is the circumradius R.
5. With O as the centre and R as the radius, draw a circle.
6. Since the distance from O to each vertex is equal to R, the circle passes through all three points A, B, and C.
7. No other circle can pass through all three points A, B, and C because the circumcentre O is unique.
8. Therefore, there is one and only one circle that passes through any three non-coll