Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

(N/A) Let $AB$ be a diameter of the circle with center $O$. Two tangents $PQ$ and $RS$ are drawn at points $A$ and $B$ respectively.
Since the radius is perpendicular to the tangent at the point of contact,we have:
$OA \perp RS$ and $OB \perp PQ$
Therefore:
$\angle OAR = 90^{\circ}$
$\angle OAS = 90^{\circ}$
$\angle OBP = 90^{\circ}$
$\angle OBQ = 90^{\circ}$
From the above,we can observe that:
$\angle OAR = \angle OBQ = 90^{\circ}$ (These are alternate interior angles)
$\angle OAS = \angle OBP = 90^{\circ}$ (These are alternate interior angles)
Since the alternate interior angles are equal,the lines $PQ$ and $RS$ must be parallel.