Question

O is the circumcentre of the triangle ABC and OD is perpendicular on BC, Prove that $\angle BOD=\angle A$

Answer 1

Given : O is the circumcentre of $\Delta ABC$

$OD\perp BC$

OB is joined

To prove : $\angle BOD=\angle A$

Construction : Join OC.

Proof : Arc BC subtends $\angle BOC$ at the centre and $\angle BAC$ at the remaining part at the circle.

$\therefore \angle BOC=2\angle A................\left(i\right)$

In right $\Delta OBD$ and $\Delta OCD$

Side OD = OD (Common)

Hyp. OB = OC (Radii of the circle)

$\therefore \Delta OBD=\Delta OCD$ (RHS crierion)

$\therefore BOD=\angle COD=\frac{1}{2}\angle BOC$

$\Rightarrow \angle BOC=2\angle BOD$ ......(ii)

From (i) and (ii)

2 $\angle BOD=2\angle A$

$\therefore \angle BOD=\angle A$