In fig., O is the centre of the circle, PA and PB are tangent segments. Show that the quadrilateral AOBP is cyclic.
Since tangent at a point to a circle is perpendicular to the radius through the point.
∴ OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = 90° and ∠OBP = 90°
⇒ ∠OAP +∠OBP = 90° + 90° ….(i)
In quadrilateral OAPB, we have
∠OAP + ∠APB + ∠AOB + ∠OBP = 360°
⇒ (∠APB + ∠AOB) + (∠OAP +∠OBP) = 360°
⇒ ∠APB + ∠AOB + 180° = 360°
∠APB + ∠AOB = 180° ….(ii)
From equations (i) and (ii),we can say that the quadrilateral AOBP is cyclic.
Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
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