Question
Two congruent circles with centres $O$ and $O′$ intersect at two points $A$ and $B.$ Then $\angle\text{AOB}=\angle\text{AO'B}.$

Answer


Join $AB$ and $OB, O'A$ and $BO'.$
In $\triangle\text{AOB}$ and $\triangle\text{AO'B,} OA = AO' [$both circles have same radius$] OB = BO' [$both circles have same radius$]$ and $AB = AB [$common chord$]$
$\triangle\text{AOB}=\triangle\text{AO'B} [$by $SSC$ congruence rule$]$
 $\Rightarrow\angle\text{AOB}=\angle\text{AO'B}[ $by $CPCT]$

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