O is the circumcentre of the ΔABC and D is the mid-point of the base BC. Prove that ∠BOD=∠A.
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Given in a ΔABC a circle is circumscribed having centre O.
Also, D is the mid point of BC.
To prove that ∠BOD=∠Aor∠BOD=∠BAC.
Construction : Join OB, OD and OC.
OB = OC [both are the radius of circle]
BD = DC [D is the mid point of BC]
OD = OD [Common side] ΔBOD≅ΔCOD [ by SSS congruence rule] ∠BOD=∠COD [by CPCT] …..(i)
We know that in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. ∴2∠BAC=∠BOC ⇒∠BAC=12∠BOC ⇒∠BAC=12(∠BOD+∠COD) ⇒∠BAC=12×2(∠BOD) [ from Equation (i)] ⇒∠BAC=∠BOD=∠A