Q.Prove that the two tangents drawn to a circle from an external point are equal in length, and the line segments joining the points of contact to the external point subtend equal angles at the center of the circle.

Given: From an external point \(P\), two tangents \(PA\) and \(PB\) are drawn to a circle with center \(O\), where \(A\) and \(B\) are the points of contact. Lines \(OA\), \(OB\), and \(OP\) are joined, forming angles ∠POA and ∠POB at the center. To Prove: (i) \(PA = PB\) (ii) ∠POA = ∠POB Proof: Since \(PA\) and \(PB\) are tangents, and \(OA\) and \(OB\) are radii drawn to the points of contact, ∴ \(OA \perp PA\) and \(OB \perp PB\) In right-angled triangles \(POA\) and \(POB\): - ∠OAP = ∠OBP (each is a right angle) - Hypotenuse \(OP\) is common - \(OA = OB\) (radii of the same circle) ∴ ∆PAO ≅ ∆PBO [By RHS congruence criterion] ∴ \(PA = PB\) (corresponding sides of congruent triangles) — (i) Proven And ∠POA = ∠POB (corresponding angles of congruent triangles) — (ii) Proven
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