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O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that \angleBOD = \angleA.

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Given: O is the circumcentre of the triangle ABC and D is the mid-point of the base BC.

\therefore OD is perpendicular BC

In the right triangles OBD and OCD.

We have OB = OC                  (Radii of the same circle)

OD = OD                                 (common)

\angleODB = \angleODC                     (90o)

\therefore \triangleOBD \cong \triangleOCD                  (R.H.S. congruence)

\Rightarrow \angleBOD = \angleCOD                 (CPCT)

\Rightarrow \angle BOD =\frac{1}{2} \angle BOC            … (1)

Also, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle. Consider arc BC,

\Rightarrow \angle BAC =\frac{1}{2} \angle BOC           … (2)

From (1) and (2), we have

\therefore  \angleBOD = \angleBAC

Hence Proved.

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