Question

$O$ is the centroid of $△ABC$ and $D$ is mid point of base $BC$ then prove $\angle BOD=\angle A.$

Answer 1

Given: $O$ is the circumcentre of $\Delta ABC$ and $OD\perp BC$

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

Construction: Join $OB$ and $OC$

Proof: In $\Delta OBD$ and $\Delta OCD$, we have

$OB=OC$ [Each equal to radius of the circumcircle]

$\angle ODB=\angle ODC$ [Each of ${90}^0$]

$OD=OD$ [Common]

$\therefore \angle OBD=\angle OCD$ [By SAS congruence]

$\therefore \angle BOD=\angle COD$ [By C.P.C.T]

$\therefore \angle BOC=2\angle BOD=2\angle COD$

Now, are $BC$ subtends $\angle BOC$ at the centre and $\angle A$ at a point in the remaining part of the circle.

$\therefore \angle BOC=2\angle A$

$\Rightarrow 2\angle BOD=2\angle A[\because \angle BOC=2\angle BOD]$

$\Rightarrow \angle BOD=\angle A$

Hence proved.