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.
A O P B


Answer:


Step by Step Explanation:
  1. 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.
  2. Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
    Thus, PAOAOAP=90 and PBOBOBP=90 So, OAP+OBP=90+90=180(i)
  3. Now, the sum of all the angles of a quadrilateral is 360. Thus, AOB+OAP+APB+OBP=360AOB+APB+(OAP+OBP)=360AOB+APB=180 [Using eq(i)] 
  4. Both pairs of opposite angles have the sum 180. Thus, we can say that quadrilateral AOBP is cyclic.

You can reuse this answer
Creative Commons License