Prove that the tangents drawn at …

 
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: Since AB is a tangent to the circle at P and OP is the radius through the point of contact.
{tex}\therefore{/tex} {tex}\angle{/tex}OPA = 90^o ........ (i)
[The tangent at any point of a circle is {tex}\perp{/tex} 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 ........ (ii)
[The tangent at any point of a circle is  {tex}\perp{/tex} to the radius through the point of contact]
From eq. (i) and (ii), {tex}\angle{/tex}OPA = {tex}\angle{/tex}OQD
But these form a pair of equal alternate angles also,
{tex}\therefore{/tex} AB {tex}\parallel{/tex} CD