Question

In the given figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.

Answer 1

It is given that,∠ABC is on circumference of circle BD is passing through centre.

Construction: Join A and C to form AC and extend BO to D such that BD be the perpendicular bisector of AC.

Now in $△BDA\mathrm{and}△BDC$ we have

AD = CD (BD is the perpendicular bisector)

So $\angle BDA=\angle BDC=90°$

(Common)

$△BDA\cong △BDC\left(\mathrm{SAS}\mathrm{congruency}\mathrm{criterion}\right)$

Hence (by cpct)