Question

O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A

Answer 1

We have to prove that   ∠BOD = ∠A

Since, circumcenter is the intersection of perpendicular bisectors of each side of the triangle.

Now according to figure A, B, C are the vertices of ΔABC

In ΔBOC , OD  is perpendicular bisector of BC

So, BD = CD                 

OB = OC             (Radius of the same circle)

And,

OD = OD         (Common)

Therefore,

\[\bigtriangleup BDO \cong \bigtriangleup CDO \left( SSS \text{ congruency criterion } \right)\]

\[\angle BOD = \angle COD \left( \text{ by cpct } \right)\]

We know that angle formed any chord of the circle at the center is twice of the angle formed at the circumference by same chord

Therefore,

` angleBAC = 1/2 angleBOC`

`⇒ angleBAC = 1/2 xx 2 angleBOD`

`⇒ angleBAC = angle BOD`

Therefore,

`angleBOD = angleA`
Hence proved