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]
OP = OC [Both radii]
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.