In Figure-7, XY and MN are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and MN at B. Prove that

To prove-  AOB =
Proof-
In  AOP and  AOC,
OA =OA [Common]
AP =AC [tangents from external point A]
Therefore by SSS congruence,  AOP   AOC
and by CPCT,  PAO =  OAC
..................(i)

Similarly, from  OBC and  OBQ, we get;
QBC = 2.  OBC.............(ii)

Adding eq (1) and eq (2)

PAC +  QBC = 180
2(  OBC +  OAC) = 180
OBC +  OAC) = 90

Now, in  OAB,
Sum of interior angle is 180.
So,  OBC +  OAC +  AOB = 180
AOB = 90
hence proved.

Related Chapters

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-