Given : Circles $$C_1$$ and $$C_2$$ of radii $$r_1$$ and $$r_2$$ respectively and $$r_1 < r_2$$.
$$AB$$ and $$CD$$ are two common tangents.
To prove : $$AB = CD$$
Construction : Produce $$AB$$ and $$CD$$ upto P where both tangents meet.
Proof : Tangents from an external point to a circle are equal.
For circle $$C_1 , PB = PD$$ __(i)
and the circle $$C_2 , PA = PC$$ __(ii)
Subtracting (i) from (ii), we have
$$PA - PB = PC - PD$$
$$\Rightarrow AB = CD$$