Given: In triangle △ABC, O is the circumcenter and OD ⊥ BC. Prove that: ∠BOD = ∠BAC Let’s break it down in English: **Given:** In triangle △ABC, O is the circumcenter (the point where the perpendicular bisectors of the sides meet), and OD is perpendicular to side BC. **To Prove:** The angle ∠BOD formed at the center between points B and D is equal to the angle ∠BAC at the vertex A. This is a classic geometry result based on the properties of a circle and triangle. Would you like me to walk you through the full proof in English as well?

Q.Given: In triangle △ABC, O is the circumcenter and OD ⊥ BC. Prove that: ∠BOD = ∠BAC Let’s break it down in English: **Given:** In triangle △ABC, O is the circumcenter (the point where the perpendicular bisectors of the sides meet), and OD is perpendicular to side BC. **To Prove:** The angle ∠BOD formed at the center between points B and D is equal to the angle ∠BAC at the vertex A. This is a classic geometry result based on the properties of a circle and triangle. Would you like me to walk you through the full proof in English as well?

**Assume:** △ABC has circumcenter O, and OD ⊥ BC. **To Prove:** ∠BOD = ∠BAC **Construction:** Join B to O and C to O. **Proof:** ∠BOC is the central angle subtended by arc BC, and ∠BAC is the inscribed angle subtended by the same arc. \[ \therefore \angle BOC = 2\angle BAC \quad \text{—— (i)} \] Now, in triangles △BOD and △COD: - OB = OC (radii of the same circle) - ∠ODB = ∠ODC = 90° (since OD ⊥ BC) - OD is common \[ \therefore \triangle BOD \cong \triangle COD \Rightarrow \angle BOD = \angle COD = \frac{1}{2} \angle BOC = \frac{1}{2} \cdot 2\angle BAC = \angle BAC \quad \text{[from (i)]} \] Therefore, ∠BOD = ∠BAC — proven.