**Math Labs with Activity – Line Drawn through Centre of a Circle to Bisect a Chord**

**OBJECTIVE**

To verify that the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord

**Materials Required**

- A sheet of white paper
- A sheet of tracing paper
- A geometry box
- A piece of cardboard
- A tube of glue

**Theory**

The theorem to be verified is the converse of the theorem verified in Activity 18.

The theorem can be proved as below.

Consider a circle with centre O and radius r having a chord AB. Let M be the midpoint of the chord AB (see Figure 19.1).

Join OA, OB and OM.

In ΔAOM and BOM, we have

- OA = OB (each equal to r)
- OM = OM (common)
- AM = MB (M being the midpoint of AB)

Therefore, ΔAOM is congruent to ΔBOM (by SSS-criterion).

So, ∠OMA = ∠OMB – 90° (since ∠OMA and ∠OMB form a linear pair) i.e., OM ⊥ AB.

**Procedure**

**Step 1:** Paste the sheet of white paper on the cardboard and mark a point O on this paper. With O as the centre, draw a circle with any radius.

**Step 2:** Draw a chord AB in this circle.

**Step 3:** Trace the circle along with the chord AB on the tracing paper.

**Step 4:** Fold the tracing paper along a line which cuts the chord AB in such a way that the part of the chord that lies on one side of this line overlaps the part on the other side and the point A lies exactly over the point B.

Form a crease and unfold the tracing paper. Mark the point M where the line of fold meets the chord AB. Then, M is the midpoint of the chord AB.

**Step 5:** Join OM as shown in Figure 19.2. Then, OM is the line drawn through the centre of the circle to bisect the chord.

**Step 6:** Now, again fold the tracing paper such that . the point A lies exactly over the point B. What do you observe?

**Observations**

We observe that the fold is along the line OM. This shows that OM is perpendicular to AB. Thus, OM is the perpendicular bisector of the chord AB.

**Result**

It is verified that the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

**Remarks:**

From the above result it can be deducted that the perpendicular bisectors of two chords of a circle intersect at its centre.

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