From a point P outside the circle , with centre O, tangents PA and PB are drawn , prove that:
$∠ A O P$ = $∠ B O P$

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Solution

In $△ A O P$ and $△ B O P$ , we have
$A P = B P$ [ Tangents from P to the circle ]
$O P = O P$ [ Common ]
$O A = O B$ [ Radii of the same circle ]
Hence, by $S A S$ criterion of congruence
$△ A O P ≅△ B O P$
So, by C.P.C.T. we have
$∠ A O P$ = $∠ B O P$

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From a point P outside the circle , with centre O, tangents PA and PB are drawn , prove that: ∠AOP = ∠BOP

In △AOP and △BOP, we have AP=BP [ Tangents from P to the circle ] OP=OP [ Common ] OA=OB [ Radii of the same circle ] Hence, by SAS criterion of congruence △AOP≅△BOP So, by C.P.C.T. we have ∠AOP = ∠BOP

From a point P outside the circle , with centre O, tangents PA and PB are drawn , prove that:
$∠ A O P$ = $∠ B O P$

In $△ A O P$ and $△ B O P$ , we have
$A P = B P$ [ Tangents from P to the circle ]
$O P = O P$ [ Common ]
$O A = O B$ [ Radii of the same circle ]
Hence, by $S A S$ criterion of congruence
$△ A O P ≅△ B O P$
So, by C.P.C.T. we have
$∠ A O P$ = $∠ B O P$