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Question
In the following figure; P and Q are the points of intersection of two circles with centers O and O'. If straight lines APB and CQD are parallel to OO';
prove that: (i) OO' = `1/2`AB ; (ii) AB = CD
Solution
Drop OM and O'N perpendicular on AB and OM' and O'N' perpendicular on CD.
∴ OM, O'N, OM' and O'N' bisect AP, PB, CQ and QD respectively.
( Perpendicular is drawn from the center of a circle to a chord bisects it. )
∴ MP = `1/2"AP" , "PN" = 1/2"BP" , "M'Q" = 1/2"CQ" , "QN"' = 1/2"QD"`
Now, OO' = MN = MP + PN = `1/2( "AP + BP" ) = 1/2"AB"` ...(i)
and OO' = M'N' = M'Q + QN' = `1/2( "CQ + QD" ) = 1/2"CD"` ...(ii)
By (i) and (ii),
AB = CD.
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