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

Answer 1

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.