In Fig. 9.13, \(AB\) and \(CD\) are common tangents to two circles of unequal radii. Prove that \(AB=CD\).
\(AB=CD\)
Step 1: Let the two common tangents \(AB\) and \(CD\) meet at a point \(P\) outside the circles.
Step 2: From geometry, we know that the lengths of tangents drawn from the same external point to a circle are equal.
Step 3: Apply this property to the left circle:
Step 4: Apply the same property to the right circle:
Step 5: Look at line \(PAB\):
Step 6: Now look at line \(PCD\):
Step 7: Since \(PA = PC\) and \(PB = PD\), subtracting gives:
\[ AB = PA - PB = PC - PD = CD \]
Final Result: Hence, we have proved that \(AB = CD\).