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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 \angle AOB = 90^{\circ} 

 
 
 
 

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To prove- \angle AOB = 90^0
Proof-
In \Delta AOP and \Delta AOC,
OA =OA [Common]
OP = OC [Both radii]
AP =AC [tangents from external point A]
Therefore by SSS congruence, \Delta AOP \cong \Delta AOC
and by CPCT, \angle PAO = \angle OAC
\Rightarrow \angle PAC = 2\angle OAC ..................(i)

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

Adding eq (1) and eq (2)

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

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

 

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Safeer PP

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