Question

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Answer 1

given: In the figure O is centre of the circle AB is chord, F is a point on AB such that AB = BF

Construction: join OA = OB

Proof: In right triangle ∆ OAF and ∆ OBF

OA= OB (R) ( radii of the same circle)

OP= OP (S) (common)

therefore ∆ OAF =~ ∆ OBF ( RHS)

AP = BP (cpct)

hence proved

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