PA and PB are tangents to a circle with center O from an external point P, touching the circle at A and B respectively. Show that the quadrilateral AOBP is cyclic.
Answer:
- Given:
PA and PB are the tangents to the circle with center O from an external point P.
Here, we have to check if quadrilateral AOBP is cyclic or not.
We know that in a cyclic quadrilateral the sum of opposite angles is 180∘. - Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Thus, PA⊥OA⟹∠OAP=90∘ and PB⊥OB⟹∠OBP=90∘ So, ∠OAP+∠OBP=90∘+90∘=180∘…(i) - Now, the sum of all the angles of a quadrilateral is 360∘. Thus, ∠AOB+∠OAP+∠APB+∠OBP=360∘∠AOB+∠APB+(∠OAP+∠OBP)=360∘⟹∠AOB+∠APB=180∘ [Using eq(i)]
- Both pairs of opposite angles have the sum 180∘. Thus, we can say that quadrilateral AOBP is cyclic.