In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: “The centre of the circle lies on the angle bisector of ”.
Given that O is the centre of the circle, OA, OB, and OC are its radii. Therefore, OA = OB = OC. We are also given that chords AB and AC are congruent, so AB = AC.
Consider ΔOAB and ΔOAC:
- OA = OA (Common side)
- OB = OC (Radii of the same circle)
- AB = AC (Given)
By the SSS congruence rule, ΔOAB is congruent to ΔOAC. Since corresponding parts of congruent triangles are equal, we have ∠OAB = ∠OAC.
This implies that the line segment AO bisects ∠BAC. Hence, the centre of the circle, O, lies on the angle bisector of ∠BAC.
Explanation
The textbook context establishes that radii of a circle are equal and demonstrates the use of SSS congruence for triangles formed by two radii and a chord (as seen in the explanation for equal chords subtending equal angles at the centre). Applying this, triangles OAB and OAC have all three sides equal (two radii and the congruent chords). By CPCTC, the angles at vertex A are equal, proving AO is the angle bisector.
Solution Steps
Step 1: Identify that OA = OB = OC since they are all radii of the same circle.
Step 2: Note that AB = AC (given, as the chords are congruent).
Step 3: Apply SSS congruence to show ΔOAB ≅ ΔOAC.
Step 4: Conclude ∠OAB = ∠OAC by CPCTC, meaning AO bisects ∠BAC.
Step 5: State that since O lies on AO, O lies on the angle bisector of ∠BAC.