Question

Question 7
O is the circumcentre of the $\Delta ABC$ and D is the mid-point of the base BC. Prove that $\angle BOD=\angle A$.

Answer 1

Given in a $\Delta ABC$ a circle is circumscribed having centre O.

Also, D is the mid point of BC.
To prove that $\angle BOD=\angle Aor\angle BOD=\angle BAC$.
Construction : Join OB, OD and OC.

Proof
In $\Delta BODand\Delta COD$
OB = OC [both are the radius of circle]
BD = DC [D is the mid point of BC]
OD = OD [Common side]
$\Delta BOD\cong \Delta COD$ [ by SSS congruence rule]
$\angle BOD=\angle COD$ [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.
$\therefore 2\angle BAC=\angle BOC$
$\Rightarrow \angle BAC=\frac{1}{2}\angle BOC$
$\Rightarrow \angle BAC=\frac{1}{2}(\angle BOD+\angle COD)$
$\Rightarrow \angle BAC=\frac{1}{2}\times 2\left(\angle BOD\right)$ [ from Equation (i)]
$\Rightarrow \angle BAC=\angle BOD=\angle A$
Hence proved.