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

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 $\u2206$ AOC and $\u2206$ BOD

OA = OB ( Given )

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

And

OC = OD ( Given )

Thus ,

$\u2206$ AOC $\cong $ $\u2206$ BOD ( By SAS rule )

Therefore,

**AC = BD ( By CPCT ) ( Hence proved )**

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