$O$ is the circumcentre of the triangle $ABC$ and $OD$ is perpendicular on $BC$. Prove that $\angle BOD = \angle A$.
Given:
$O$ is the circumcentre of the triangle $ABC$ and $OD$ is perpendicular on $BC$.
To do:
We have to prove that $\angle BOD = \angle A$.
Solution:
Join $OC$.
Arc $BC$ subtends $\angle BOC$ at the centre and $\angle BAC$ at the remaining part of the circle.
Therefore,
$\angle BOC = 2\angle A$....…(i)
In right angled triangle $OBD$ and $\triangle OCD$
$OD = OD$ (Common)
$OB = OC$ (Radii of the circle)
Therefore, by RHS congruence,
$\triangle OBD \cong \triangle OCD$
This implies,
$\angle BOD = \angle COD = \frac{1}{2} \angle BOC$
$\angle BOC = 2\angle BOD$.........…(ii)
From (i) and (ii), we get,
$2\angle BOD = 2\angle A$
$\angle BOD = \angle A$
Hence proved.
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