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

Given: From an external point \(P\) to a circle with center \(O\), two tangents \(PA\) and \(PB\) are drawn, touching the circle at points \(A\) and \(B\) respectively. The points \(O, A, B\) and \(P\) are joined such that the segments \(PA\) and \(PB\) form angles \(\angle POA\) and \(\angle POB\) at the center. To prove: (i) \(PA = PB\) (ii) \(\angle POA = \angle POB\) Proof: \(PA\) 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 \(POB\): \(\angle OAP = \angle OBP\) (each is a right angle) The hypotenuse \(OP\) is common And \(OA = OB\) (radii of the same circle) ∴ \(\triangle POA \cong \triangle POB\) (By RHS congruence rule) Therefore, \(PA = PB\) (corresponding sides of congruent triangles) — (i) Proved \(\angle POA = \angle POB\) (corresponding angles of congruent triangles) — (ii) Proved
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