If \(AB\) and \(CD\) are common tangents to two two circles of equal radii, prove that \(AB = CD\).
Proof: \(AB\) and \(CD\) are common tangents to two circles of equal radii.
Join \(OA\), \(OC\), \(O'B\) and \(O'D\).
Tangent at any point on a circle is perpendicular to the radius through the point of contact.
\(OA \perp AB\) and \(OC \perp CD\)
\(\angle OAB = 90^\circ\) and \(\angle OCD = 90^\circ\)
Thus, \(AC\) is a straight line.
Also, \(\angle O'BA = \angle O'DC = \)\(^\circ\)
Thus, \(BD\) is a straight line.
Therefore, \(ABCD\) is a .
We have, \(\angle A = \angle B = \angle C = \angle D = 90^\circ\).
So, \(ABCD\) is a .
Hence, \(AB = CD\).