# pa is perpendicular to ab , qb is perpendicular to ab .and pa=qb.prove triangle oap congruent to triangle obq.is oa =op

From given information we form our diagram , as , and we take point  " O "  where line PB and QA intersect .

Now In $∆$ PAB and $∆$QBA

PA  =  QB                                            ( Given )

$\angle$ PAB  =  $\angle$QBA  = 90$°$              ( Given )
And
AB =  AB                                           (  Common side )

Hence

$∆$ PAB $\cong$$∆$QBA                          ( By SAS rule )
So,

$\angle$ APB  =  $\angle$ BQA                        (  By CPCT )                                                 --------------- ( 1 )

And
PB  =  QA                                          (  By CPCT )                                                 --------------- ( 2 )

Now In $∆$ OAP and $∆$OBQ

PA  =  QB                                            ( Given )

$\angle$ APO  =  $\angle$BQO                          (  From equation 1 ) (  Here $\angle$ APB   and  $\angle$ APO are same angles and $\angle$ BQA  and  $\angle$ BQO    are same angles )
And
$\angle$ AOP  =  $\angle$ BOQ                         (  Vertically opposite angles )
Hence

$∆$ OAP $\cong$$∆$OBQ                          ( By AAS rule )                                              ( Hence proved )
So,

OA =  OB                                            (  By CPCT )                                                 --------------- ( 3 )

OP =  OQ                                            (  By CPCT )                                                 --------------- ( 4 )

So. from equation 2 ,3 and 4 , we get

OA  =  OP                                          ( Hence proved )

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