Given: In triangle ABC, O is the circumcenter and OD ⊥ BC. Prove: \(\angle\)BOD = \(\angle\)BAC.

Q.Given: In triangle ABC, O is the circumcenter and OD ⊥ BC. Prove: \(\angle\)BOD = \(\angle\)BAC.

Assume that \( \triangle ABC \) has circumcenter O and OD ⊥ BC. Construction: Join O and C. Proof: In triangles \( \triangle BOD \) and \( \triangle COD \): - OB = OC (both are radii of the circle) - ∠ODB = ∠ODC = 90° - OD is common ∴ \( \triangle BOD \cong \triangle COD \) ⇒ ∠BOD = ∠COD i.e., ∠BOD = \( \cfrac{1}{2} \angle BOC \) …(i) Now, for arc BC, ∠BOC is the central angle and ∠BAC is the inscribed angle. ∴ ∠BAC = \( \cfrac{1}{2} \angle BOC \) …(ii) From equations (i) and (ii): ∠BOD = ∠BAC [Proved]