$AB$ and $CD$ are common tangents to two circles. If the radii of the two circles are equal,prove that $AB = CD$.

(N/A) Given: $AB$ and $CD$ are common tangents to two circles with centers $O$ and $O^{\prime}$ and equal radii $r$.
To prove: $AB = CD$.
Construction: Join $OA, OC, O^{\prime}B$ and $O^{\prime}D$.
Proof:
$1$. Since $AB$ is a tangent to the circle at $A$,the radius $OA$ is perpendicular to $AB$. Thus,$\angle OAB = 90^{\circ}$.
$2$. Similarly,since $CD$ is a tangent to the circle at $C$,the radius $OC$ is perpendicular to $CD$. Thus,$\angle OCD = 90^{\circ}$.
$3$. Since $AB$ and $CD$ are parallel tangents (as they are perpendicular to the line joining the centers if the radii are equal),$AC$ is a diameter or a line segment perpendicular to the tangents.
$4$. In the quadrilateral $ABDC$,we have $\angle A = 90^{\circ}$,$\angle B = 90^{\circ}$,$\angle C = 90^{\circ}$,and $\angle D = 90^{\circ}$.
$5$. Since the radii are equal,the distance between the parallel tangents $AB$ and $CD$ is constant,making $AC = BD = 2r$.
$6$. $A$ quadrilateral with all angles equal to $90^{\circ}$ and opposite sides equal is a rectangle.
$7$. Therefore,$ABDC$ is a rectangle.
$8$. Hence,the opposite sides are equal,which implies $AB = CD$.