In the following figure: P and Q are the points of intersection of two circles with centres O and O’. If straight lines APB and CQD are parallel to OO’. Prove that
(i) OO’ = ½ AB
(ii) AB = CD

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Solution

Draw OM and O'N are perpendicular on AB and OM’ and O’N’ are perpendicular on CD.

So, OM, O’N, OM’ and O’N’ bisect AP, PB, CQ and QD respectively

[Perpendicular drawn from the centre of a circle to a chord bisects it.]

MP = ½ AP, PN = ½ BP, M’Q = ½ CQ, QN’ = ½ QD

We know that

OO’ = MN = MP + PN = ½ (AP + BP) = ½ AB ……. (i)

OO’ = M’N’ = M’Q + QN’ = ½ (CQ + QD) = ½ CD …… (ii)

Equating (i) and (ii)

AB = CD

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In the following figure: P and Q are the points of intersection of two circles with centres O and O’. If straight lines APB and CQD are parallel to OO’. Prove that (i) OO’ = ½ AB (ii) AB = CD

In the following figure: P and Q are the points of intersection of two circles with centres O and O’. If straight lines APB and CQD are parallel to OO’. Prove that
(i) OO’ = ½ AB
(ii) AB = CD