In Question 5 above, if the radii of the two circles are equal, prove that \(AB=CD\).
\(AB=CD\) (the tangents are parallel in this case).
Step 1: We are given two circles of equal radius. Let the radius of each circle be \(r\) (in SI units, radius is measured in metres, \(m\)).
Step 2: Draw their common external tangents. These tangents will touch the first circle at points \(A\) and \(B\), and the second circle at points \(C\) and \(D\).
Step 3: Because the radii of the circles are equal, the distance of the centre of each circle from any tangent is the same, equal to \(r\).
Step 4: This means that the two external tangents are parallel lines. Why? Because each tangent is at the same distance from both centres.
Step 5: Now consider the two parallel tangents. The line joining the centres of the circles will cut both tangents at right angles. This line works like a transversal crossing two parallel lines.
Step 6: By symmetry, the distance between the points of contact on the first tangent (segment \(AB\)) is equal to the distance between the points of contact on the second tangent (segment \(CD\)).
Step 7: Therefore, we conclude that \(AB = CD\).
Note: This result can also be seen as a special case of Question 5 when the external centre of similarity goes to infinity.