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  • Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD

Answers (1)

Given: Two equal chords AB and CD of a circle intersect at a point P.

To Prove: PB = PD

Proof: Given,

Join OP draw OL \perp AB and OM \perp CD

AB = CD

OL = OM                                (equal chords are equidistant from the centre)

Consider \triangleOLP and \triangleOMP

OL = OM                               

\angleOLP = \angleOMP                      (both are 90o)

OP = OP

\triangleOLP \cong \triangleOMP                      (SAS congruence)

LP = MP                     …(i)    (CPCT)

AB = CD                                 (Given)

\frac{1}{2}\left ( AB \right )=\frac{1}{2}\left ( CD \right )

BL = DM                    …(ii)

Subtracting equation (ii) from equation (i) we get

LP – BL = MP – DM

\Rightarrow PB = PD

Hence proved.

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