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.

Given:

Two equal chords of a circle intersect within the circle

To do:

We have to prove that the segments of one chord are equal to corresponding segments of the other chord.

Solution:

Let $AB$ and $CD$ be two equal cords which intersect at point $R$.

From the centre of the circle, draw a perpendicular to $AB$ and another one perpendicular to $CD$

$OP \perp AB$

$OQ \perp CD$.

Join $OR$.

From the diagram,

$OP$ bisects $AB$ and $OP \perp AB$

$OQ$ bisects $CD$ and $OQ \perp CD$

$AB = CD$

This implies,

$AP = QD$.........(i)

$PB = CQ$.........(ii)

In triangles $OPR$ and $OQR$,

$\angle OPR = \angle OQR$

$OR = OR$ (Common side)

$OP = OQ$ ($AB$ and $CD$ are equal and they are equidistant from the centre)

Therefore, by RHS congruency,

$\triangle OPR \cong \triangle OQR$

This implies,

$PR = QR$..........(iii) (CPCT)

From (i) and (ii), we get,

$AP+PR = QD+QR$

$AR = RD$

From (ii) and (iii), we get,

$PB-PR = CQ-QR$

$BR = CR$

Hence proved.

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