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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.