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In following figure from exterior point ‘P’ two tangent line PA and PB touches the circle at A and B, then prove that PAOB is a cyclic quadrilateral where O is the center of circle. |
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Answer» We know that tangent at any point of a circle is perpendicular to radius through the point of contact. OA ⊥ PA and OB ⊥ PB ∴ ∠OAP = ∠OBP = 90° ⇒ ∠OAP + ∠OBP = 90° + 90° = 180° In quadrilateral PAOB ∠PAO + ∠AOB + ∠OBP + ∠BPA = 360° ⇒ ∠AOB + ∠BPA = 360° – 180° ⇒ ∠AOB + ∠BPA=180° ∵ Sum of opposite angle of quadrilateral is = 180°. ∴ Quadrilateral PAOB is a cyclic quadrilateral |
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