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Question
In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD, prove that AB = CD.
In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD, prove that AB = CD.
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Solution
Given: BC is a diameter of a circle with centre O. AB and CD are two chords such that AB || CD.
TO prove: AB = CD
Construction: Draw OL ⊥ AB and OM ⊥ CD.
Proof:
In Δ OLB and Δ OMC, we have:
∠ OLB = ∠ OMC
[90° each]
∠ OBL = ∠ OCD
[Alternate angles as AB || CD ]
OB = OC
[Radii of a circle]
∴ Δ OLB ≅ Δ OMC (AAS criterion)
Thus, OL = OM (CPCT)
We know that chords equidistant from the centre are equal.
Hence, AB = CD
Given: BC is a diameter of a circle with centre O. AB and CD are two chords such that AB || CD.
TO prove: AB = CD
Construction: Draw OL ⊥ AB and OM ⊥ CD.
Proof:
In Δ OLB and Δ OMC, we have:
∠ OLB = ∠ OMC
[90° each]
∠ OBL = ∠ OCD
[Alternate angles as AB || CD ]
OB = OC
[Radii of a circle]
∴ Δ OLB ≅ Δ OMC (AAS criterion)
Thus, OL = OM (CPCT)
We know that chords equidistant from the centre are equal.
Hence, AB = CD
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