students must be able to understand why ∠ at Centre = 2 times ∠ at Circumference.[br][br][br]Proof:[br]Let ∠AOC = 2a[br]Let ∠BOC = 2b[br]Then ∠AOB = 360° - 2a – 2b[br]∠ OCA = 90° – a (isosceles triangle)[br]∠BCO = 90° – b (isosceles triangle)[br]Therefore, ∠ACB = (90° – a) + (90° – b) = 180° – a – b [br]Hence, ∠AOB = 2∠ACB (∠ at Centre = 2 times ∠ at Circumference) Proven[br][br]http://weelookang.blogspot.sg/2014/12/geogebra-angle-at-centre-equal-twice.html

Steps:[br][br]1. Compare angles at the centre of a circle with angle touching the circumference.[br]2. vary the ∠ at Centre O for which it is acute less than 90 °[br]3. write down the value of ∠ at Centre O and ∠ at Circumference point A[br]4. vary the ∠ at Centre O for which it is obtuse more than 90° and less than 180°.[br]5. do step 3[br]6. vary the ∠ at Centre O for which it is reflex more than 180°.[br]7. do step 3[br][br]Thinking:[br][br]looking at the evidence of the table of recorded values, suggest a relationship between [br]∠ at Centre O and ∠ at Circumference point A.[br][br][br]Conclusion:[br][br]∠ at Centre = 2 times ∠ at Circumference.