O is the ex-centre of the ∆ A B C . Prove that ∠ B O C = 90 ° - 1 2 ∠ B A C .

For the ∆ A B C , O is the ex-centre. We need to prove, ∠ B O C = 90 ° - 1 2 ∠ B A C . Construction: Join O B and O C . Extend the lines A B and A C up to the lines such that they meet at G and H respectively. We know that in a triangle, the exterior angle is equal to the sum of two interior opposite angles. ∠ G B D = ∠ B A D + ∠ A D B . . . . . . . . . . . . . . i ∠ D C H = ∠ C A D + ∠ A D C . . . . . . . . . . . . . . ii We know that ∠ D B O = 1 2 ∠ G B D and ∠ D C O = 1 2 ∠ D C H From i and ii , we can write, ∠ D B O = 1 2 ∠ B A D + 1 2 ∠ A D B and ∠ D C O = 1 2 ∠ C A D + 1 2 ∠ A D C In ∆ B O C , ∠ B O C + 1 2 ∠ C A D + 1 2 ∠ A D C + 1 2 ∠ B A D + 1 2 ∠ A D B = 180 ° . . . . . . . . . . . . . . . iii ⇒ ∠ B O C + 1 2 ∠ C A D + 1 2 ∠ B A D + 1 2 × 180 ° = 180 ° ⇒ ∠ B O C + 1 2 ∠ B A C + 90 ° = 180 ° ⇒ ∠ B O C = 90 ° - 1 2 ∠ B A C Hence, proved.