Prove tangent drawn at the end …

Given: PQ is a diameter of a circle with centre O. The lines AB and CD are the tangents at P and Q respectively.
 
To prove: AB {tex}\parallel{/tex} CD
Proof: AB is a tangent to the circle at P and Op is the radius through the point of contact
{tex}\because{/tex} {tex}\angle{/tex}OPA = 90^o .......(1) [The tangent at any point of a circle is perpendicular to the radius through the point of contact]
{tex}\because{/tex} CD is a tangent to the circle at Q and OQ is the radius through the point of contact.
{tex}\therefore{/tex} {tex}\angle{/tex}OQD = 90^o ........(2) [The tangent at any point of a circle is perpendicular to the radius through the point of contact]
From (1) and (2),
{tex}\angle{/tex}OPA  = {tex}\angle{/tex}OQD
But these form a pair of equal alternate angles.
{tex}\because{/tex} AB {tex}\parallel{/tex} CD