In the given figure, $P A$ and $P B$ are two tangents drawn from an external point $P$ to a circle with centre $O$ . Prove that $O P$ is the right bisector of line segment $A B$ .

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Solution

Join $O A$ & $O B$

In $Δ O A P$ & $Δ O B P$

$∠ O A P = ∠ O B P = 90^{o}$
$O P = O P$
$O A = O B$ (radii)
$∴ Δ O A P ≅ Δ O B P$ by RHS
Hence $∠ B P O = ∠ A P O$ (by cpct)

In $Δ A P X$ & $Δ B P X$
$A P = B P$ (Tangents from same pt)
$∠ A P X = ∠ B P X$ (proved above)
$P X = P X$
$∴ Δ A P X ≅ Δ B P X$ by SAS criterion

Thus $A X = B X$ & $∠ A X P = ∠ B X P$
$\Rightarrow P X ⊥ A B$ .

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From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that :

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