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 = $\dfrac{AB}{2}$ and CN = DN = $\dfrac{CD}{2}$

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.

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