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

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Solution

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 $△ B D A and △ B D C$ we have

AD = CD (BD is the perpendicular bisector)

So $∠ B D A = ∠ B D C = 90 °$

(Common)

$△ B D A ≅ △ B D C (SAS congruency criterion)$

Hence (by cpct)

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