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In the figure, $O$ is the centre of the circle, $BO$ is the bisector of $\angle ABC$. Show that $AB = BC$.
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Given:

$O$ is the centre of the circle, $BO$ is the bisector of $\angle ABC$. 

To do:

We have to show that $AB = BC$.

Solution:

Draw $OL \perp AB$ and $OM \perp BC$


In $\triangle OLB$ and $\triangle OMB$,

$\angle 1 = \angle 2$              (Given)

$\angle L = \angle M= 90^o$

$OB = OB$                           (Common side)

Therefore, by AAS axiom,

$\triangle OLB \cong \triangle OMB$

This implies,

$OL = OM$                  (CPCT)

$OL$ and $OM$ are distances from the centre and chords equidistant from the centre are equal.

$BA = BC$

Hence proved.

Updated on: 10-Oct-2022

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