Given :
∆ABC whose circumcenter is point O and D is mid point of BC
∴ BD = DC
To Proved :
∠BOD = ∠A
Construction: Draw OD ⊥ BC and join OB and OC
In ∆OBD and ∆OCD
OB = OC (radius of same circle)
∠ODB = ∠ODC (each 90°)
OD = OD (common)
∴ ∆OBD = ∆OCD
⇒ ∠BOD = ∠COD
⇒ ∠BOC = 2∠BOD = 2∠COD
∵ Angle subtended by arc BC at the center of circle
i.e., ∠BOC is double the angle subtended by same arc at point A of remaining part of circle
i.e., ∠BAC.
∠BOC = 2∠A
⇒ 2∠BOD = 2∠A [∵∠BOC = 2∠BOD]
⇒ ∠BOD = ∠A
