8.1.12

12 ABC is a triangle in inscribed in a circle whose centre is O and OD is perpendicular drawn O to BC prove ∠BOD ≡ ∠BAC.

Given:  : In ΔOBD and ΔOCD

               Since OD ⏊BC

               BD  =   CD

               OD = OD (∵ common side)

               BO = CO( ∵ radii )

               ∴ Δ BOD ≡ Δ COD (∵ SSS )

              ∠ OBD = ∠ COD

             ∴ ∠ BOD = ½ ∠ COD ......Eq₁

Again,

              ∠A = ½ ∠BOC ( ∵ Theorem 1 ).....Eq₂

From Eq₁ and Eq₂ ,

              ∠ BAC = ∠ BOC