Question

Please solve question number 6

Answer 1

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Given ;- AB= CD

and intersect at C

=> draw two perpendicular As OM ⊥CD and ON ⊥ AB 

=> AN  =  NB   = AB2
And
CM  = MD = CD2

And we know AB  =  CD (  Given )  , So 

AN  =  NB  =  CM  =  MD                                                  ------------- ( 1 )

Now In  ∆ ONE and ∆ OME 

ON =  OM                                                    

     ( As we know Equal chords of a circle are equidistance from the center  )

OE  =  OE                                                          ( Common side )
And
∠ ONE =  ∠ OME   = 90°               

( From construction )

Hence

∆ ONE ≅ ∆ OME                                    


    ( by RHL rule  )
So,
NE  =  ME                                                     

     (  By CPCT )                  ------------- ( 2 )

SO,

AE  =  AN  +  NE    , CE  =  CM    +  ME
And
EB  =  NB  -  NE  ,  ED  =  MD  -  ME 
So,
From equation 1 and 2 , we get

AE  =  CE                                                                                              
   ------------- (  3 )

And
EB  =  ED                                                                                                 ------------- ( 4 )

Now In  ∆ AED and ∆ CEB

AE  =  CE                                                         ( From equation 3   )

EB =  ED                                                          ( From equation 4 )
And
∠ AED =  ∠ CEB                            

     ( Vertically opposite angles  )

Hence

∆ AED ≅ ∆ CEB                                 

       ( By SAS rule )

So,

AD  =  CB                                                 

       (  From CPCT )                                                                          ( Hence proved )

$hope \: helps$