In the given figure, XY and XY are two parallel tangents to a circle with center O and another tangent AB with the point of contact C intersects XY at A and XY at B. Prove that AOB=90.

O Q P C B A X X' Y' Y


Answer:


Step by Step Explanation:
  1. We know that tangent at any point is perpendicular to the radius through the point of contact.  So, APQ=90 and BQP=90.(i)
  2. Consider quadrilateral APQB APQ+BQP+QBA+PAB=360[Sum of angles of a quadrilateral is 360.]QBA+PAB=360(APQ+BQP)QBA+PAB=360(90+90)[Using eq (i)]QBA+PAB=180QBC+PAC=180 [As, PAB is same as PAC and  QBA is same as QBC] (ii)
  3. We know that the tangents from an external point are equally inclined to the line segment joining the center to that point.

    So, CAO=12PAC and CBO=12QBC Therefore, CBO+CAO=12(QBC+PAC)CBO+CAO=12×180=90 [Using eq (ii)] ABO+BAO=90[As,CAO is same as BAO and CBO is same as ABO] (iii)
  4. In AOB, we have BAO+AOB+ABO=180[Sum of angles of a triangle is 180.]AOB=180(ABO+BAO)AOB=18090=90[Using eq (iii)] AOB=90

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