XY an X'Y' ARE 2 parallel …

Given: In figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B.
To Prove : {tex}\angle{/tex}AOB = 90^o
Construction: Join OC

Proof: {tex}\angle{/tex}OPA = 90^o ........ (i)
{tex}\angle{/tex}OCA = 90^o........ (ii)
[Tangent at any point of a    circle is ⊥ to  the radius through the point of contact]
In right angled triangles OPA and OCA,
OA = OA [Common]
AP = AC [Tangents from an external
point to a circle are equal]
{tex}\therefore{/tex} {tex}\triangle{/tex}OPA {tex}\cong{/tex} {tex}\triangle{/tex}OCA [RHS congruence criterion]
{tex}\therefore{/tex} {tex}\angle{/tex}OAP = {tex}\angle{/tex}OAC [By C.P.C.T.]
{tex}\Rightarrow{/tex} {tex}\angle{/tex}OAC = {tex}\frac12{/tex}{tex}\angle{/tex}PAB ....... (iii)
Similarly, {tex}\angle{/tex}OBQ = {tex}\angle{/tex}OBC
{tex}\Rightarrow{/tex} {tex}\angle{/tex}OBC = {tex}\frac12{/tex}{tex}\angle{/tex}QBA  .......... (iv)
{tex}\because{/tex} XY ∥ X'Y' and a transversal AB intersects them.
{tex}\therefore{/tex} {tex}\angle{/tex}PAB + {tex}\angle{/tex}QBA = 180^o [Sum of the consecutive interior angles on the same side
of the transversal is 180^o
{tex}\Rightarrow{/tex} {tex}\frac12{/tex}{tex}\angle{/tex}PAB + {tex}\frac12{/tex}{tex}\angle{/tex}QBA = {tex}\frac12{/tex}{tex}\times{/tex} 180^o.......... (v)
{tex}\Rightarrow{/tex} {tex}\angle{/tex}OAC + {tex}\angle{/tex}OBC = 90^o [From eq. (iii) & (iv)]
In {tex}\triangle{/tex}AOB,
{tex}\angle{/tex}OAC + {tex}\angle{/tex}OBC + {tex}\angle{/tex}AOB = 180^o [Angel sum property of a triangle]
{tex}\Rightarrow{/tex} 90^o + {tex}\angle{/tex}AOB = 180^o [From eq. (v)]