Question

Answer the last (village well) question fast pls

Answer 1

hi dear here is the answer

From given information we form our diagram , As  :

And we know  " The tangent at any point of a circle is perpendicular to the radius through the point of contact. "

So ,

∠ OPA  =  ∠ ORA  =  ∠ OQB  =  ∠ ORB  =   90°

Now , In ∆ OPA and ∆ ORA 

OP  =  OR                                                      (  Radius of circle )

∠ OPA  =  ∠ ORA  =   90°                       ( From the property of tangent )

And

OA  =  OA                                                   ( Common side ) ( Hypotenuse )

Hence ,
∆ OPA ≅ ∆ ORA                                  ( by RHS congruency rule   )

So,

AP  =  AR                                                   ( By CPCT )                                               --------------- ( 1 )

Similarly In ∆ OQB and ∆ ORB , we get  

BQ  =  BR                                                                                                                      ----------------- ( 2 )

Now we apply Pythagoras theorem In  In ∆OPT  and get

OT2  =  TP2  +  OP2

OT2 = ( TA  +  AP )2 +  OP2                                                                                   --------------------- ( 3 )
 
And , we apply Pythagoras theorem In  In ∆OQT  and get

OT2  =  TQ2  +  OQ2

OT2 = ( TB  +  BQ )2 +  OQ2                                                                                   --------------------- ( 4 )

From equation 3 and 4 , we get 

( TA  +  AP )2 +  OP2   = ( TB  +  BQ )2 +  OQ2    

Here OP =  OQ  =  radius of circle , SO  

⇒( TA  +  AP )2 +  OP2   = ( TB  +  BQ )2 +  OP2    

⇒( TA  +  AP )2   = ( TB  +  BQ )2

Now we substitute values from equation 1 and 2 and get

⇒( TA  +  AR )2   = ( TB  +  BR )2

Taking square root on  both hand side , we get

⇒TA  +  AR  =   TB  +  BR                                                                                                         ( Hence proved )

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