 Hello and welcome to the session. The given question says, proof that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle and using the above to the following in figure 3 where this is figure 3. A, B and C are three points on the circle with center O such that angle A O B is 90 degrees, A O C is 110 degrees we have to find angle B A C. First let us prove the theorem. Here we are given circle with center O, subtending angle the center and angle A C B at any point. So here we have two cases and for both the cases we shall show that the center at the angle A O B is twice the angle subtended by the same arc circle that is angle A construction S equals, so this implies angle to angle A C O and here in this case angle O A equal to angle O A C since is equal to angle two interior opposite angles and this is the property of a triangle equal to angle O C A in both the cases therefore in place of angle O A C we can write A C O or O C A. So here also in place of O A C we shall write angle O C A and this gives angle A O B is equal to two times of angle O C equation number one times of angle B C O let this be equation number two. Now to prove the first case we shall add equation one and two. So on the left hand side we have angle A O B plus angle B O P is equal to twice angle O C A plus angle B C O B O P gives angle C O gives angle equation two minus equation one O is this angle and we have to subtract A from it. So we get angle B C subtended by the same arc at the remaining part of the circle and by using this theorem let us equal to 160 degrees. Now we know that from the above theorem that the angle subtended equal to twice angle B A is 160 degrees so this is equal to two times of angle B A to 160 degrees divided by two which gives hope you have understood it by and take care.