Math Proof: Angle at circumference is twice angle at centre

A mini project by JH301 Group C^^

1. How do we prove that the angle at the centre is twice the angle at the circumference?

2. O P B C A Y With the given diagram, we draw a line OP and a line OC , where AC is the diameter of the circle. Since,

3. O P B C A Y We know that AO = OC = OP = OB, because the radii of a circle are all equal. We can then identify all the isosceles triangles. Also,

4. Now, we are ready to solve!

5. O B C A Y Let’s look at  AOB.  AOB lies on the diameter of the circle. Therefore, we can say that  AOB = 180- (  COP +  POB). We call that equation 1. 1:  AOB = 180 – (  COP +  POB) P

6. O B C A Y P 1:  AOB = 180 – (  COP +  POB) 2: POB = 180 - 2 OPB Now let’s look at angle POB. Angle POB is part of an isosceles triangle POB, where OPB = OBP . Therefore, we can say that POB = 180 – (OPB+ OBP) Or POB = 180 - 2 OPB

7. 1:  AOB = 180 – (  COP +  POB ) 2: POB = 180 - 2 OPB Given the 2 equations, we’ll substitute equation 2 into equation 1. AOB = 180 – (  COP + 180 - 2 OPB ) = 180 – 180 - COP + 2 OPB = 2  OPB - COP And… TADA!!! We have our 3 rd equation formed!

8. O B C A Y P Getting back to our circle, We know that OPB = OPA + APB Simple. No explanation needed:D And that will be equation 4! 3: AOB = 2 OPB - COP 4: OPB = OPA + APB

9. 3: AOB = 2 OPB - COP 4: OPB = OPA + APB Of course then, the next step will be to substitute 4 into 3! AOB = 2 (OPA + APB ) - COP = 2 OPA + 2 APB - COP  2 OPA + 2 APB - COP + AOB = 0 And here, we have our equation 5!!! 5: 2 OPA + 2 APB - COP + AOB = 0 Alright, back to the circle.

10. O B C A Y P We know that AOP is an isoceles triangle where OAP = OPA Therefore, we can say that AOP + 2 OPA = 180 However, we also know that AOP + COP = 180 (Angles on a straight line) Thus, we can see that 2 OPA = COP (Which will be equation 6!) *This is also a property of triangles, where the sum of 2 interior opposite s = the exterior . 5: 2 OPA + 2 APB - COP + AOB = 0 6: 2 OPA = COP

11. 5: 2 OPA + 2 APB - COP + AOB = 0 6: 2 OPA = COP Again, we substitute equation 6 into 5. 2 OPA + 2 APB - 2 OPA + AOB = 0 2 APB - AOB = 0 2 APB = AOB And…..

12. VIOLA!!!!

13. We’ve proven that AOB is twice APB , where  AOB is an  at the centre, and  APB is an  at the circumference. Conclusion: Angle at the centre is twice angle at the circumference. O B C A Y P

14. A simple mathematical proof brought to you by JH301 Math Group C Credits: Mr Christopher Cheng Miss Maureen Ng