In the figure, O is the centre of the circle, BO is the bisector of $∠ A B C$ . Show that AB = BC.

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Solution

Given: In the figure, a circle with centre O, OB is the bisector of $∠ A B C$

To prove : AB = BC

Construction : Draw OL $⊥$ AB and OM $⊥$ BC

Proof : In $Δ O L B a n d Δ O M B$ ,

$∠ 1 = ∠ 2$ (Given)

$∠ L = ∠ M$ $(E a c h = 90^{\circ})$

$O B = O B$ (Common)

$∴ Δ O L B ≅ Δ O M B$ (AAS criterion)

$∴ O L = O M$ (c.p.c.t)

But these are distance from the center and chords equidistant from the centre are equal

$∴$ Chord $B A = B C$

Hence, $A B = B C$

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