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प्रश्न
Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD.
उत्तर
Given: Two equal chords AB and CD of a circle intersecting at a point P.
To prove: PB = PD
Construction: Join OP, draw OL ⊥ AB and OM ⊥ CD
Proof: We have, AB = CD
⇒ OL = OM ...[Equal chords are equidistant from the centre]
In ΔOLP and ΔOMP,
OL = OM ...[Proved above]
∠OLP = ∠OMP ...[Each 90°]
And OP = OP ...[Common side]
∴ ΔOLP ≅ ΔOMP ...[By RHS congruence rule]
⇒ LP = MP [By CPCT] ...(i)
Now, AB = CD
⇒ `1/2 (AB) = 1/2 (CD)` ...[Dividing both sides by 2]
⇒ BL = DM ...(ii) [Perpendicular drawn from centre to the circle bisects the chord i.e., AL = LB and CM = MD]
On subtracting equation (ii) and equation (i), we get
LP – BL = MP – DM
⇒ PB = PD
Hence proved.
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