Q.Prove that the two tangents drawn from an external point to a circle are equal in length from the point to the points of contact on the circle.

Given: From an external point \(P\), two tangents \(PA\) and \(PB\) are drawn to a circle with center \(O\), touching the circle at points \(A\) and \(B\) respectively. Lines \(OA\), \(OB\), and \(OP\) are joined, forming angles ∠POA and ∠POB at the center. To Prove: \(PA = PB\) Proof: Since \(PA\) and \(PB\) are tangents, and \(OA\) and \(OB\) are radii drawn to the points of contact, \[ \Rightarrow OA \perp PA \quad \text{and} \quad OB \perp PB \] In right-angled triangles △POA and △POB: - ∠OAP = ∠OBP = 90° - Hypotenuse \(OP\) is common - \(OA = OB\) (radii of the same circle) \[ \Rightarrow △POA \cong △POB \quad \text{[By RHS congruence criterion]} \Rightarrow PA = PB \quad \text{[Corresponding sides of congruent triangles]} \]
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