If O is the circumcentre of Triangle ABC and OD prependicular to BC. Prove that
Angle BOD = Angle A
O is the circumcentre of ∆ABC and OD⊥BC.
∴ O is the point of intersection of perpendicular bisectors of the sides of ∆ABC.
⇒ D is the mid point of BC.
⇒ BD = DC
In ∆OBD and ∆OCD,
OB = OC (Radius of the circle)
OD = OD (Common)
BD = DC (Proved)
∴ ∆OBD ≅ ∆OCD (SSS Congruence criterion)
⇒∠BOD = ∠COD (CPCT)
We know that, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠BOC = 2 ∠BAC
⇒ 2∠BOD = 2 ∠A [ Using (1) ]
⇒ ∠BOD = ∠A
look angle BOD = 1/2 OF angle BOC (as OD IS THE PERPENDICULAR to BC )...(1)
angle A = 1/2 OF Angle BOC (angle subtended at the centre by the chord is double than the angle subtended by it at any other point on the remaining part of the circle) ..........(2)
from equation 1 and 2 it is clear that angle BOD = angle A.
HOPE IT HELPS YOU