# Question 4 (ii)

Dear Student,

As here we want to solve only second part of query ( 4 ) , So we assume that first part's solution you already have ( As that information needed in solution for second part ) , Then

From first part we know :  OO' = $\frac{1}{2}$AB                                 --- ( 1 )

And our diagram , As :

Here we draw two perpendiculars from O to CQ and O' to DQ

And we know if a perpendicular is drawn to any chord from center of circle then that perpendicular also bisect that chord , So

CM =  QM  = $\frac{1}{2}$ CQ                                        --- ( 2 )

And

DN =  QN  = $\frac{1}{2}$ DQ                                        --- ( 3 )

And

MN  =  QM  +  QN , Substitute values from equation 2 and 3 we get :

MN  = $\frac{1}{2}$ CQ +  $\frac{1}{2}$ DQ ,

MN  =  $\frac{1}{2}$ ( CQ + DQ ) ,

MN = $\frac{1}{2}$ CD

And here we also know :  Line CQD | | OO' and from construction we know OM $\perp$ CQ and O'N $\perp$ DQ , So we can say that

OMNO' is a rectangle and in rectangle opposite sides are equation in length , So  MN =  OO' that we substitute in above equation we get :

OO' = $\frac{1}{2}$ CD   , And from equation 1 we get
$\frac{1}{2}$ AB = $\frac{1}{2}$ CD  ,

AB  =  CD                                                                   ( Hence proved )