In the figure, \(AB\) is a tangent touching the circle at \(C\) and cuts two parallel tangents \(XY\) and \(X′Y′\) at \(A\) and \(B\). If \(O\) is the centre of the circle. Prove that \(∠AOB\) is a right angle
Proof:
Taking, \(∆OAP\) and \(∆OAC\) in consideration \(OP = OC\) [\(OP\) and \(OC\) radius of the circle]
\(OA=\)
We know that, The tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(∠OPA = ∠OCA =\) \(^°\)
Therefore by congruency, \(∆OAP ≅ ∆OAC\)
So, \(∠POA = ∠\)
we can conclude that, \(∠COP =\) ----- (1)
Similarly, \(∠COQ =\) ------(2)
Adding (1) and (2), \(RHS = LHS\) we get,
\(2 (∠COA + ∠COB) =\) \(^°\)
\(2 (∠\)\() =\) \(^°\)
\(∠AOB = 90^°\).
Hence, proved.