OA = OB, OC = OD, Angle AOB = Angle COD. Prove that AC=BD
 

Dear Student,

Please find below the solution to the asked query:

Given : $\angle$ AOB =  $\angle$ COD                              --- ( 1 )

And

$\angle$ COB  =  $\angle$ COB                                          --- ( 2 )     (  Same angles ) 

So, If we subtract equation 2 in equation 1 we get  :

$\angle$ AOB - $\angle$ COB =  $\angle$ COD - $\angle$ COB

We know " Euclid's third Axiom: If equals be subtracted from equals, the remainders are equal. "  we get :

$\angle$ AOC  =  $\angle$ BOD                                          --- ( 3 )

In $∆$ AOC and $∆$ BOD 

OA  =  OB                                                      ( Given )

$\angle$ AOC  =  $\angle$ BOD                                      ( From equation 3 )

And

OC  =  OD                                                      ( Given )

Thus ,

$∆$ AOC $\cong$ $∆$ BOD                                        ( By SAS rule )

Therefore,

AC  =  BD                                                      ( By CPCT )                           ( Hence proved )


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