Question

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

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  $$△BDA\text{ and }△BDC$$ we have

AD = CD           (BD is the perpendicular bisector) 

So ,

$$\mathrm{\angle }BDA=\mathrm{\angle }BDC=90°$$

BD = BD  (Common)

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

Hence AB =BC   (by cpct)