If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
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Question

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

KnowledgeBoat · Accepted Answer

Let AB and CD be the two equal chords (AB = CD = a). Let the chords intersect at point P. Join OP. Draw OM and ON perpendicular to chords AB and CD respectively. We know that, Perpendicular from center bisects the chord. ∴ AM = MB = and CN = DN = Since, AB = CD. ∴ AM = MB = CN = DN = x(let) .....(1) In ∆ OMP and ∆ ONP, ⇒ ∠M = ∠N (Both equal to 90°) ⇒ OP = OP (Common side) ⇒ OM = ON (Equal chords are equidistant from the center.) ∴ ∆ OMP ≅ ∆ ONP (By R.H.S congruence rule) We know that, Corresponding parts of congruent triangles are equal. ∴ MP = NP = y(let) (By C.P.C.T.) .....(2) From figure, ⇒ CP = CN - NP = x - y and PB = MB - MP = x - y ∴ CP = PB. From figure, DP = CD - CP = a - (x - y) and AP = AB - BP = a - (x - y) ∴ AP = PD. Hence, proved that if two equal chords of a circle intersect within the circle, the segments of one chord are equal to corresponding segments of the other chord.

Mathematics

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Circles

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