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O is the circumcentre of the triangle ABC and D is mid-point of the base BC. Prove that ∠BOD = ∠A.

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Answer

From the below figure:

O is the circumcentre of the triangle ABC and D is mid-point of the base BC. Prove that ∠BOD = ∠A. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Arc BC subtends ∠BOC at center and ∠BAC at the point A on the circle.

∴ ∠BOC = 2∠A

In △OBD and △ODC,

OD = OD (Common side)

BD = CD (As D is the mid-point of BC)

OB = OC (Radius of the same circle)

∴ △OBD ≅ △ODC (SSS rule of congruency).

∴ ∠BOD = ∠COD (As corresponding part of congruent triangles are congruent.)

Since, ∠BOD = ∠COD so,

∠BOD = 12\dfrac{1}{2}∠BOC ….(i)

∠BOC = 2∠A
∠A = 12\dfrac{1}{2} ∠BOC …..(ii)

From (i) and (ii) we get,

∠BOD = ∠A

Hence, proved that ∠BOD = ∠A.

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